# Java.lang.StrictMath.expm1() Method

## Description

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.

## Declaration

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

```public static double expm1(double x)
```

## Parameters

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

## Example

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
```
java_lang_strictmath.htm