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Solve over Domains

Sandu Ursu

Posted 7 years ago

To my knowledge, $ \mathbb{R} \subset \mathbb{C}$ and not vice-versa. Then I don't understand why I get more real-valued solutions with Reals, than with Complexes: $Assumptions = a > 0 && a \[Element] Reals && b \[Element] Reals && t \[Element] Reals && t >= 0 && c1 \[Element] Reals && c2 \[Element] Reals && c \[Element] Reals && B \[Element] Reals && c2 > 0 && c1 < 0 Solve[Log[Abs[(B - c2)/(B - c1)]] == (a t - c) (c2 - c1), B, Reals] // Simplify Solve[Log[Abs[(B - c2)/(B - c1)]] == (a t - c) (c2 - c1), B, Complexes] // Simplify

To my knowledge, $ \mathbb{R} \subset \mathbb{C}$ and not vice-versa.

Then I don't understand why I get more real-valued solutions with Reals, than with Complexes:

$Assumptions = 
 a > 0 && a \[Element] Reals && b \[Element] Reals && 
  t \[Element] Reals && t >= 0 && c1 \[Element] Reals && 
  c2 \[Element] Reals && c \[Element] Reals && B \[Element] Reals && 
  c2 > 0 && c1 < 0

Solve[Log[Abs[(B - c2)/(B - c1)]] == (a t - c) (c2 - c1), B, Reals] // Simplify

Solve[Log[Abs[(B - c2)/(B - c1)]] == (a t - c) (c2 - c1), B, Complexes] // Simplify

POSTED BY: Sandu Ursu

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 7 years ago

In cases like these, Mathematica ought to issue a warning that the solution set is not complete.

POSTED BY: Gianluca Gorni

Sandu Ursu

Posted 7 years ago

Should I view this as a bug or as a property?

POSTED BY: Sandu Ursu

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 7 years ago

I think it is an intrinsic limitation of `Solve`, which is meant mainly for polynomial or analytic functions, and `Abs` is not analytic. The solution set of Abs[z]==1 in the complex plane cannot be expressed as simple substitution rules. The set of solutions to `Abs[z]==1` is handled correctly by `Reduce` in an implicit way, though.