3 ms·
expm1 avoids catastrophic cancellation (due to finite precision floating point), and so is more accurate for x close to zero than naive(x) = exp(x) - 1, e.g. ex
by dbaupp 8y ago
expm1 avoids catastrophic cancellation (due to finite precision floating point), and so is more accurate for x close to zero than naive(x) = exp(x) - 1, e.g. expm1(1e-100) != 0, but naive(1e-100) = 0.