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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 4

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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}]]
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  • $\begingroup$ That totally makes sense now +1. Thanks a lot! $\endgroup$
    – Remi.b
    Commented Mar 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}}

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{Solve [ x^2 == 1, {x}], Solve [ x^2 == 1 && x > 0, {x}]}
(* {{{x -> -1}, {x -> 1}}, {{x -> 1}}}*)
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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.

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