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I use Reduce to find conditions on some parameters such that certain inequalities are fulfilled. An additional condition is that all parameters are positive. The first problem is that I have to mention all those conditions when calling Reduce, and the second is that they appear in the result and make it less readable.

I tried to use Assumptions, but it doesn't change the result.

I wonder if it is possible to hide these trivial conditions or at least suppress them in the output?

$Assumptions = Flatten[{Thread[{r, p, h, K} > 0], Element[{r, p, h, K}, Reals]}]
Reduce[K*(r - h)/r > 0 && (p*h)/(r - h) < p && K >0 && p>0 && h>0 && r>0,h]
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    $\begingroup$ MemberQ[Keys[Options[Reduce]], Assumptions] returns False, so it is inappropriate to use $Assumptions with Reduce[]. $\endgroup$ Commented Jan 22, 2020 at 9:23
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    $\begingroup$ Dmitry, apply Simplify to the result of Reduce. $\endgroup$ Commented Jan 22, 2020 at 9:24
  • $\begingroup$ @HenrikSchumacher, this seems to be the most simple and efficient solution so far. Thank you! $\endgroup$ Commented Jan 22, 2020 at 9:38
  • $\begingroup$ You're welcome! $\endgroup$ Commented Jan 22, 2020 at 9:45

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Here is one possibility:

conds = K > 0 && p > 0 && h > 0 && r > 0;
Complement[Reduce[K*(r - h)/r > 0 && (p*h)/(r - h) < p && conds, h, Reals], conds]
   0 < h < r/2
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  • $\begingroup$ Unfortunately, it does not suppress the conditions from the output when inequalities get more complex, e.g., here: conds = K > 0 && p > 0 && h > 0 && r > 0 && [Rho] > 0; Complement[Reduce[K*(r - h)/r > 0 && (p*h)/(r - [Rho] - h) < p && conds, h, Reals], conds] $\endgroup$ Commented Jan 22, 2020 at 9:35
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    $\begingroup$ Then you can use Simplify[], which does take assumptions: With[{conds = K > 0 && p > 0 && h > 0 && r > 0 && ρ > 0}, Assuming[conds, Simplify[Reduce[K*(r - h)/r > 0 && (p*h)/(r - ρ - h) < p && conds, h, Reals]]]]. $\endgroup$ Commented Jan 22, 2020 at 9:42

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