0
$\begingroup$

I have a generated a data set of a function $\alpha(\lambda_2,\varepsilon):\mathbb{R}_+ \times (0,1) \mapsto (0,1)$. I used mathematica to generate a list by varying $\lambda_2$ and $\varepsilon$; $\varepsilon$ is very small, and $\lambda$ is in between 0.5 and 1.0.

{Nλ1, Nμ1, Nμ2, NCN} = {0., 1., 1., 7060};
Clear[Tableϵ, Tableλ, Tableαϵλ];
Tableϵ = Table[10^-m, {m, 3, 15, 1/3}];
Tableλ = Table[m, {m, 0.5, 1.0, 1/20}];
Tableαϵλ = {};

For[j = 1, j <= Length[Tableλ], j++, {
    lastkϵ = NCN - 1;
    For[i = 1, i <= Length[Tableϵ], i++, {
        With[{Nλ2 = Tableλ[[j]], ϵ = Tableϵ[[i]]}, {
            For[k = lastkϵ, k > Nλ2 NCN, k--, {
                If[ε[Nλ1, Nλ2, Nμ1, Nμ2, k/NCN, NCN] < ϵ && Nλ2 < k/NCN < 2 Nλ2, {
            AppendTo[Tableαϵλ, {ϵ, Nλ2, N[k/NCN]}],
            Break[]
                }];
            }]
        }];
    lastkϵ = k;
    }];
}];

Now, I'd like to generate some nice plots. In fact, I wanted to see the curves of $\alpha \approx k/NCN$ as a function of $\varepsilon$ using $\lambda_2$ as series (eventually excluding some to look nicer.

My question is, how do I break this data set into series that I can filter for some plots?

Thank you in advance.

Δ1[λ1_, λ2_, μ1_, μ2_, c0_ ] :=  1/4 + (μ2/(λ1 + λ2) (c0 μ1 - λ1 - λ2 )/( 2 μ1 - μ2));
Sp1[λ1_, λ2_, μ1_, μ2_, c0_ ] := 1/2 + Sqrt[Δ1[λ1, λ2, μ1, μ2, c0]]; Sp2[λ1_, λ2_, μ1_, μ2_, c0_ ] := 1/2 - Sqrt[Δ1[λ1, λ2, μ1, μ2, c0]];
Δ2[λ1_, λ2_, μ1_, μ2_, c0_ ] := (2 λ1 μ1 + 2 λ2 μ1 - λ1 μ2 - λ2 μ2)^2 -     3 (2 λ2 μ1 - λ2 μ2) (-2 λ1 μ1 - 2 λ2 μ1 + λ1 μ2 + λ2 μ2); Δ3[λ1_, λ2_, μ1_, μ2_, c0_ ] :=  2 (2 λ1 μ1 + 2 λ2 μ1 - λ1 μ2 - λ2 μ2)^3 - 9 (2 λ2 μ1 - λ2 μ2) (2 λ1 μ1 + 2 λ2 μ1 - λ1 μ2 - λ2 μ2) (-2 λ1 μ1 - 2 λ2 μ1 + λ1 μ2 + λ2 μ2) + 27 (2 λ2 μ1 - λ2 μ2)^2 (λ1 μ2 + λ2 μ2 - c0 μ1 μ2);  
Sq1[λ1_, λ2_, μ1_, μ2_, c0_ ] := -1/(3 λ2) (λ1 + λ2 + 2 Sqrt[Δ2[λ1, λ2, μ1, μ2, c0]] Cos[1/3 ArcTan[Sqrt[4 Δ2[λ1, λ2, μ1, μ2, c0]^3/Δ3[λ1, λ2, μ1, μ2, c0]^2 - 1]] + (2 π )/3]); 
Sq2[λ1_, λ2_, μ1_, μ2_, c0_ ] := -1/(3 λ2) (λ1 + λ2 + 2 Sqrt[Δ2[λ1, λ2, μ1, μ2, c0]] Cos[1/3 ArcTan[Sqrt[4 Δ2[λ1, λ2, μ1, μ2, c0]^3/Δ3[λ1, λ2, μ1, μ2, c0]^2 - 1]] + (4 π )/3]); 
Sq3[λ1_, λ2_, μ1_, μ2_, c0_ ] := -1/(3 λ2) (λ1 + λ2 + 2 Sqrt[Δ2[λ1, λ2, μ1, μ2, c0]] Cos[1/3 ArcTan[Sqrt[4 Δ2[λ1, λ2, μ1, μ2, c0]^3/Δ3[λ1, λ2, μ1, μ2, c0]^2 - 1]]]);
F[λ1_, λ2_, μ1_, μ2_, c0_, n_] := 1 - (2 λ2 μ1 + λ1 μ2 - c0 μ1 μ2 )/(λ2 (2 μ1 - μ2)) (    1 - Sq3[λ1, λ2, μ1, μ2, c0]/Sp1[λ1, λ2, μ1, μ2, c0]  )/(1 - Sq3[λ1, λ2, μ1, μ2, c0]) Sp1[λ1, λ2, μ1, μ2, c0]^-n  
ε[λ1_, λ2_, μ1_, μ2_, c0_, CN_] := Chop[(2 λ2 μ1 + λ1 μ2 - c0 μ1 μ2)/(λ2 (2 μ1 - μ2)) (1 - Sq3[λ1, λ2, μ1, μ2, c0]/Sp1[λ1, λ2, μ1, μ2, c0])/(1 - Sq3[λ1, λ2, μ1, μ2, c0]) Sp1[λ1, λ2, μ1, μ2, c0]^Round[-CN (1 - c0)] ] 
πminus[λ1_, λ2_, μ1_, μ2_, c0_] := (c0 - (λ1 + λ2)/μ1)/(λ2 (2/μ2 - 1/μ1))
|improve this question
$\endgroup$
  • 1
    $\begingroup$ By using those symbol names you're looking for trouble! Defne your symbols starting with a lowercase letter $\endgroup$ – Dr. belisarius Mar 22 '16 at 13:43
  • $\begingroup$ there is no alpha in your code.. so its not clear what you are asking. You probably just need do use Select to pull out what you want from the list though. $\endgroup$ – george2079 Mar 22 '16 at 13:55
  • $\begingroup$ Below the code, I say $\alpha \approx k / NCN$. $\endgroup$ – Guilherme Thompson Mar 22 '16 at 14:01
  • $\begingroup$ right so something like ListPlot[{#[[3]],[[1]]}&/@Select[Tableael,#[[2]]==val&]] $\endgroup$ – george2079 Mar 22 '16 at 14:23
  • 1
    $\begingroup$ Dear Guilherme, I urge you to read this: mathematica.stackexchange.com/q/7924/8 $\endgroup$ – Verbeia Mar 22 '16 at 21:35
2
$\begingroup$ 
ListPlot[Select[
  Table[#[[{3, 1}]] & /@ 
    Select[Table\[Alpha]\[Epsilon]\[Lambda], #[[2]] == lam & ],
   {lam, Table\[Lambda]}], Length[#] > 0 &], Joined -> True, 
 PlotRange -> All]

enter image description here

|improve this answer
$\endgroup$

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.