(expm1 [x])Returns e^x - 1. Near 0, expm1(x)+1 is more accurate to e^x than exp(x).
If x is ##NaN => ##NaN
If x is ##Inf => #Inf
If x is ##-Inf => -1.0
If x is zero => x
See: https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#expm1-double-
Examples
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