expm1(x) will be more accurate than exp(x) - 1 for x << 1. Decimal version of std.math function. Reference: Beebe, Nelson H. F., "Computation of expm1(x) = exp(x) - 1".
See Implementation
expm1(x) will be more accurate than exp(x) - 1 for x << 1. Decimal version of std.math function. Reference: Beebe, Nelson H. F., "Computation of expm1(x) = exp(x) - 1".