Java - Math expm1(double) method

---

---

Description

The Java Math expm1(double x) 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).Special 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.

The computed result must be within 1 ulp of the exact result. Results must be semi-monotonic. The result of expm1 for any finite input must be greater than or equal to -1.0. Note that once the exact result of e^x - 1 is within 1/2 ulp of the limit value -1, -1.0 should be returned.

Declaration

Following is the declaration for Java Math expm1() method

public static double expm1(double x)

Parameters

x − the exponent to raise e to in the computation of e^x -1.

Return Value

This method returns the value e^x - 1.

Exception

NA

Example 1

The following example shows the usage of Math expm1() method.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = 0.5;
   
      // print e raised at x
      System.out.println("Math.expm1(" + x + ")=" + Math.expm1(x));
   }
}

Output

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

Math.expm1(0.5)=0.6487212707001282

Example 2

The following example shows the another usage of Math expm1() method with zero value.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = 0;
   
      // print e raised at x
      System.out.println("Math.expm1(" + x + ")=" + Math.expm1(x));
   }
}

Output

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

Math.expm1(0.0)=0.0

Example 3

The following example shows the usage of Math expm1() method with one value.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = 1;
   
      // print e raised at x
      System.out.println("Math.expm1(" + x + ")=" + Math.expm1(x));
   }
}

Output

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

Math.expm1(1.0)=1.718281828459045