Something changed with Solve
between versions 12.1 and 12.2.
12.1:
Solve[n == n E^(r (1 - n)), n]
(* Solve::ifun -- Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. *)
(* {{n -> 0}, {n -> 1}} *)
$Assumptions = {n >= 0};
Solve[n == n E^(r (1 - n)), n]
(* Solve::ifun -- Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. *)
(* {{n -> 0}, {n -> 1}} *)
12.2:
Solve[n == n E^(r (1 - n)), n]
(* Solve::ifun -- Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. *)
(* {{n -> 0}, {n -> 1}} *)
$Assumptions = {n >= 0};
Solve[n == n E^(r (1 - n)), n]
Is this an improvement or a bug? It seems hard for the condition ConditionalExpression[1, Re[r] == 0 || Re[r] > 0 || Re[r] < 0]
not to hold, but maybe I'm overlooking something.
Two work-arounds:
$Assumptions = {n >= 0, r \[Element] Reals};
Solve[n == n E^(r (1 - n)), n]
Solve[n == n E^(r (1 - n)), n, Reals]
both give {{n -> 0}, {n -> 1}}
(no Solve::ifun
either).
Another example, not fixable by including Reals
:
$Assumptions = {n1 >= 0, n2 >= 0};
Solve[0 == n1 (1 - n1 - 0.5 n2), n1]
To be precise, my version is
$Version
(* 12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020) *)
Update:
So, I think I'll just use Assumptions->{}
to avoid these conditionals for now (while keeping $Assumptions
set for use in Simplify
).
Solve
(updated) — now takesAssumptions
options". So there must be many chances under the hood. $\endgroup$