I used Bob Hanlon's solution from here: How do I know which is the positive root for this quadratic?

or the following function but it outputs nothing, not even an empty set. It shows its been run but nothing outputs, why?

f[i_] := 1/(
   b2 b5 (b1 + i β) ϵ2 (μ + i β σ))
    i (-b1 (b2 b3 (b4 b5 - 
           a3 q3 γ δ) - γ (b2 b5 q2 + 
           a2 b2 q3 δ + b5 q1 ϵ1 + 
           a1 q3 δ ϵ1) ϵ2) (μ + 
        i β σ) + β (ϵ1 ϵ2 (i \
γ (b5 q1 + a1 q3 δ) (μ + i β σ) + 
           A b5 (p μ + f r σ + i p β σ)) + 
        b2 (-b3 i (b4 b5 - a3 q3 γ δ) (μ + 
              i β σ) + 
           i γ (b5 q2 + a2 q3 δ) ϵ2 (μ + 
              i β σ) + 
           A b5 ϵ2 (μ - 
              p μ + (f - f r + i β - 
                 i p β) σ))));

$Assumptions = 
  q1 + q2 + q3 == 1 && a1 + a2 + a3 == 1 && And @@ Thread[{
       A, β, μ, p, r,σ, a1, a2, a3, δ, q1,
        q2, q3, b1, b2, b3, b4, b5, 
       f, ϵ1, ϵ2, γ} > 0];
sol = Solve[{f[i] == 0, i > 0}, i, Method -> Reduce] // FullSimplify
  • 2
    $\begingroup$ This evaluates for longer than I am willing to wait. Perhaps it timed out on your system. $\endgroup$
    – Bob Hanlon
    Feb 9, 2022 at 16:11
  • $\begingroup$ Both Solve and FullSimplify can be very time consuming. Rather than putting both calculations in the same evaluation, split this into two steps. Won't necessarily work, but this is the recommended approach. $\endgroup$
    – Bob Hanlon
    Feb 9, 2022 at 16:23
  • $\begingroup$ Sometimes doing Algebra I yourself is more efficient that figuring out how to get Mma to do it. Get the coefficients $a,b,c$ of your quadratic factor and plug them into the output of LogicalExpand@Reduce[(-b + Sqrt[b^2 - 4 a c])/(2 a) > 0, {a, b, c}], which gives three cases for the greater root (-b + Sqrt[b^2 - 4 a c])/(2 a) to be positive. Check the lesser root similarly, if desired. Reducing the conditions to something intelligible may be beyond the powers of human and machine, but one might try anyway. $\endgroup$
    – Michael E2
    Feb 9, 2022 at 16:42
  • $\begingroup$ BTW, using f both as an undefined parameter and a defined function, while seeming not to cause trouble here, is still asking for trouble. It would be better to use a longer name for f[i], for instance. $\endgroup$
    – Michael E2
    Feb 9, 2022 at 16:44


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