public static native double 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.

`x` | the exponent to raise e to in the computation of
e^{ x} -1. |

*e*^{ x} - 1.

Diagram: Math