0
$\begingroup$

I am currently trying to obtain an analytical expression for a range of validity of an expression. I am doing this with reduce in mathematica but it keeps giving me the expression in terms of the Roots function. I tried to solve it and these roots have analytical solutions, is there any way to get a cleaner expression, preferably in terms of the parameters only and no root function... Thank you

Code:

 FullSimplify[Reduce[{-((6 \[Alpha]^4 + 21 \[Alpha]^3 \[Lambda] + 27 \[Alpha]^2 \[Lambda]^2 + 15 \[Alpha] \[Lambda]^3 + 3 \[Lambda]^4 + Sqrt[3] Sqrt[-(\[Alpha] + \[Lambda])^6 (-72 + 16 \[Alpha]^3 \[Lambda] + 21 \[Lambda]^2 + 4 \[Alpha] \[Lambda] (-9 + 4 \[Lambda]^2) + 4 \[Alpha]^2 (-15 + 8 \[Lambda]^2))])/( 4 (\[Alpha] + \[Lambda])^4)) < 0 && -((6 \[Alpha]^4 + 21 \[Alpha]^3 \[Lambda] + 27 \[Alpha]^2 \[Lambda]^2 + 15 \[Alpha] \[Lambda]^3 +  3 \[Lambda]^4 - Sqrt[3] Sqrt[-(\[Alpha] + \[Lambda])^6 (-72 + 16 \[Alpha]^3 \[Lambda] + 21 \[Lambda]^2 + 4 \[Alpha] \[Lambda] (-9 + 4 \[Lambda]^2) +4 \[Alpha]^2 (-15 + 8 \[Lambda]^2))])/( 4 (\[Alpha] + \[Lambda])^4)) < 0 && -((3 (-2 \[Beta] + \[Lambda] + \[Lambda] \[Mu]))/(\[Alpha] \
+ \[Lambda])) < 0 && \[Alpha] (\[Alpha] + \[Lambda]) > -3 && \[Lambda] \
(\[Lambda] + \[Alpha]) > 3}, {\[Alpha], \[Beta], \[Lambda], \[Mu]}, 
Reals, Cubics -> True, Quartics -> True]]

enter image description here

enter image description here

$\endgroup$
2
$\begingroup$

To convert low order Root objects use ToRadicals

sol1 = FullSimplify[
   Reduce[-((6 α^4 + 21 α^3 λ + 
           27 α^2 λ^2 + 15 α λ^3 + 
           3 λ^4 + 
           Sqrt[3] Sqrt[-(α + λ)^6 (-72 + 
                16 α^3 λ + 21 λ^2 + 
                4 α λ (-9 + 4 λ^2) + 
                4 α^2 (-15 + 
                   8 λ^2))])/(4 (α + λ)^4)) < 
      0 && -((6 α^4 + 21 α^3 λ + 
           27 α^2 λ^2 + 15 α λ^3 + 
           3 λ^4 - 
           Sqrt[3] Sqrt[-(α + λ)^6 (-72 + 
                16 α^3 λ + 21 λ^2 + 
                4 α λ (-9 + 4 λ^2) + 
                4 α^2 (-15 + 
                   8 λ^2))])/(4 (α + λ)^4)) < 
      0 && -((3 (-2 β + λ + λ μ))/(α + λ)) < 
      0 && α (α + λ) > -3 && λ (λ + α) > 3, {α, β, λ, μ}, Reals]] // ToRadicals

(* long output removed *)

Further simplification requires additional assumptions/constraints on the parameters, e.g., if all the parameters are positive

sol2 = Simplify[sol1, Thread[{α, β, λ, μ} > 0]]

(* 2 β < λ + λ μ && ((λ <= (1/(
      48 α))(-21 - 
        32 α^2 + (441 + 3072 α^2 + 256 α^4)/(-9261 + 
           152064 α^2 + 84096 α^4 + 4096 α^6 + 
           288 Sqrt[
            2] α Sqrt[(-3 + 2 α^2)^3 (1029 + 64 α^2)])^(
         1/3) + (-9261 + 152064 α^2 + 84096 α^4 + 
          4096 α^6 + 
          288 Sqrt[2] α Sqrt[(-3 + 2 α^2)^3 (1029 + 
              64 α^2)])^(1/3)) && (2 α < Sqrt[
        6] || (2 α > Sqrt[6] && 
         Sqrt[12 + α^2] < α + 2 λ))) || (Sqrt[
      12 + α^2] < α + 2 λ && 
     2 α <= Sqrt[
      6] && λ <= -(1/(
       96 α))(42 + 
         64 α^2 + ((1 - I Sqrt[3]) (441 + 3072 α^2 + 
              256 α^4))/(-9261 + 152064 α^2 + 
            84096 α^4 + 4096 α^6 + 
            288 Sqrt[
             2] α Sqrt[(-3 + 2 α^2)^3 (1029 + 64 α^2)])^(
          1/3) + (1 + I Sqrt[3]) (-9261 + 152064 α^2 + 
            84096 α^4 + 4096 α^6 + 
            288 Sqrt[
             2] α Sqrt[(-3 + 2 α^2)^3 (1029 + 64 α^2)])^(
          1/3)))) *)

Note that the expression contains complex subexpressions, e.g., (1+I*Sqrt[3]).

$\endgroup$
  • 1
    $\begingroup$ Also there is no guarantee that the parametrized Root objects will correspond to particular radical variants for all values of the parameters (there can be crossovers). This might or might not matter for a particular application. $\endgroup$ – Daniel Lichtblau Jul 16 '18 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.