I am trying to use Mathematica to quickly determine the sign of a derivative. When I use the function Sign, Mathematica informs me the answer depends on the sign of a numerator, which is rather long. Following some advice on using $Assumptions, I have now tried the following, but without the desired effect; I would have thought that Mathematica would have replaced Sign[ ] with -1 or 1 when it has been input the restrictions:
$Assumptions = 0 <= a <= 1 && 0 <= b <= 1 && c == 1
Sign[(b c (-I \[Pi] + Log[-b + (-1 + b) a] - Log[-b + (1 + b) a]) + c (-4 Log[a] + I (-2 + b) (\[Pi] + I Log[-b + (-1 + b) a]) + (2 + b) Log[-b + (1 + b) a]) a)/(b^2 (-1 + a)^3)]
Sign[0]
is0
,Sign[Indeterminate]
isIndeterminate
, andSign[z]
isz/Abs[z]
ifz
is a nonzero complex number. That last two can occur in the specified domain. $\endgroup$$Assumptions
(such asSimplify
) in order for the assumptions to have an effect. I don't thinkSign
uses$Assumptions
. $\endgroup$