1
$\begingroup$

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)]
Improve this question
$\endgroup$
3
  • $\begingroup$ Is my code correct? $\endgroup$
    – user120911
    Aug 2, 2018 at 15:29
  • 1
    $\begingroup$ The Sign[0] is 0, Sign[Indeterminate] is Indeterminate, and Sign[z] is z/Abs[z] if z is a nonzero complex number. That last two can occur in the specified domain. $\endgroup$
    – Michael E2
    Aug 3, 2018 at 12:30
  • 1
    $\begingroup$ Also, generally you have to apply a function that makes use of $Assumptions (such as Simplify) in order for the assumptions to have an effect. I don't think Sign uses $Assumptions. $\endgroup$
    – Michael E2
    Aug 3, 2018 at 12:32

1 Answer 1

Reset to default
1
$\begingroup$

You could use $Assumptions=... to define global attributes of variables!

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.