Comment #7 on issue 3266 by [email protected]: Numeric issues in hyperbolic
functions
http://code.google.com/p/v8/issues/detail?id=3266
Oh, on #4, perhaps atanh(x) = log1p(x)/2-log1p(-x)/2 is better from an
accuracy viewpoint than 1/2*log1p(-2*x/(1+x)). Also atanh(x) can't overflow
for any value in it's domain except |x| = 1, so might as well do
(log1p(x)-log1p(-x))/2 and save one division (multiplication).
But in any case, using log1p in any form is better than direct evaluation
from the definition.
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