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Compute log(1+x) accurately for small values of x



y = log1p(x) computes log(1+x), compensating for the roundoff in 1+x. log1p(x) is more accurate than log(1+x) for small values of x. For small x, log1p(x) is approximately x, whereas log(1+x) can be zero.

See Also

log, expm1

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