# Tag Info

5

As the old documentation states: $EqualTolerance gives the number of decimal digits by which two numbers can disagree and still be considered equal according to Equal. The default setting is equal to Log[10, 2^7], corresponding to a tolerance of 7 binary digits. On my system$MachinePrecision is ~15.9546 which means there are 53 bits: ...

4

UPDATE A more accurate explanation than the culprit being "low variability" is that because all of the dependent variable values begin with "-0.907" or "-0.908" essentially "eats up"/"wastes" the first 3 significant digits. Simply subtracting the minimimum value of the dependent variable works even better than standardizing by the mean and standard ...

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