Reduce[
Abs[-((4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] +
Abs[-((4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] < 1 , Abs[p]]
It is taking lot of time. It is running. Can any one help to reduce the inequality?
Answer
Because the question seeks an expression for the modulus of p
, it makes sense to express p
and q
in terms of the moduli and phases.
sim = Simplify[(Abs[-((4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] +
Abs[-((4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2)]) /.
{p -> pm Exp[I pp], q -> qm Exp[I qp]}, pm >= 0 && qm >= 0 && (pp | qp) ∈ Reals]
(* (4 (pm + qm))/Abs[-1 + Sqrt[1 + 4 E^(I pp) pm + 4 E^(I qp) qm]]^2 *)
In what follows, we explore sim <= 1
instead of the question's sim < 1
in order to obtain solutions at the boundary, sim == 1
, which is where most solutions seem to lie. Although
Reduce[sim <= 1 && pm >= 0 && qm >= 0, pm]
still produced no answer, even after 19 hours, the special case of setting qp
to π
did. Some hand-holding was required, however.
Reduce[(sim /. {qp -> Pi}) <= 1 && 2 π > pp >= 0 && pm >= 0 && qm >= 0, pm]
returned unevaluated with the message
Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>
However,
Reduce[FullSimplify[Reduce[(sim /. {qp -> Pi}) <= 1 && pm >= 0 && qm >= 0, pm],
2 Pi > pp >= 0 && pm >= 0 && qm >= 0] && 2 Pi > pp >= 0 && pm >= 0 && qm >= 0, pm]
did produce a meaningful answer.
(* (0 <= pp < 2 π && qm >= 1/4 && pm == 0) || (pp == π && qm >= 1/4 && pm >= 0) ||
(pp == π && ((qm > 1/4 && pm >= 0) || (0 <= qm <= 1/4 && pm >= 1/4 (1 - 4 qm)))) *)
Note that, except for the solution pm == 0
, all these solutions require pp == π
.
In summary, solutions are available for qp -> π
and perhaps other cases. Whether a solution can be obtained in general within several hours of computation is unknown.
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