I am trying to solve an equation by assuming that all the variables are real and strictly positive. I can use the keyword Reals for the argument dom in the Solve function Solve[expr,vars,dom]. Is there an equivalent for strictly positive Reals? Something like PosReals
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4 Answers
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1
There's a misunderstanding here. The third "dom" argument is not simply a set over which we solve the equation. There are only a few choices that can be used for the domain argument, and they have very specific effects on how Solve works. An example from the documentation:
If dom is Reals, or a subset such as Integers or Rationals, then all constants and function values are also restricted to be real.
So you can't use e.g.
Solve[x^2 == 1, x, Interval[{0, Infinity}]]
The proper way to do this, as @belisarius said, is to append the constraint to the system of equations:
Solve[x^2 == 1 && x > 0, x]
In version 10 we can also do
Solve[x^2 == 1 && x ∈ Interval[{0, Infinity}], x]
or even
Solve[x^2 == 1, x ∈ Interval[{0, Infinity}]]
answered Mar 27, 2015 at 0:39
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$\begingroup$ That totally makes sense now +1. Thanks a lot! $\endgroup$– Remi.bMar 27, 2015 at 0:45
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New in Mathematica 12 is PositiveReals (and others like NonNegativeIntegers, etc):
Solve[x^2 == 1, x, PositiveReals]
{{x -> 1}}
answered Jun 18, 2019 at 20:31
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{Solve [ x^2 == 1, {x}], Solve [ x^2 == 1 && x > 0, {x}]}
(* {{{x -> -1}, {x -> 1}}, {{x -> 1}}}*)
answered Mar 27, 2015 at 0:30
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If you are dealing with multiple variables, I suggest you to use Assumptions under function Refine, like below
Refine[Reduce[{52*n^2 + 19*n*nb - 42*n*nf > 0, 52*n^2 - 61*n*nb - 42*n*nf > 0}, {n, nb, nf}], Assumptions -> n > 0 && nb > 0 && nf > 0]
This feature can be used for most type operations. For reference, please see Assumptions.
