Tag Info

New answers tagged

0

Clear[log1p] log1p[x_, n : _Integer?Positive : 2] := (Series[Log[1 + y], {y, 0, n}] // Normal) /. y -> x log1p[1.0*^-15] 9.999999999999995*^-16


3

Problem The problem with Log[1. + 1.*^-15] not yielding 1. is not due to Log, but to MachinePrecision inputs, which I think the OP implied in the question statement: 1 + 1.*^-15 % - 1 (* 1. 1.11022*10^-15 *) So Log[1 + 1.*^-15] does return the right answer, 1.11022*10^-15, for the actual input. Solution Here is a simple way to get log1p-type ...



Top 50 recent answers are included