Compute log(1+x) accurately for small values of x
y = log1p(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.