# 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() method ` `import` `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
```

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