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I want to define a piecewise function, which has two arguments, one of which is a function of the other's piecewise argument. I am giving an example

X0 = 1.7635;
X1 = 4.4855;
γ[En_] := En/0.13957
β[En_] := Sqrt[γ[En]^2 - 1]/γ[En]
X[En_] := Log[10, β[En]*γ[En]]
f[En_, Evaluate[X[En_] _]] :=Piecewise[{{0, X[En] < X0}, {4*X[En], X[En] > X0}}]
PLot[f[En, Evaluate[X[En]]], {En, 0.001, 1000}]

Is it possible?

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Why not defining f[En_] :=Piecewise[{{0, X[En] < X0}, {4, X[En] > X0}}] ? – b.gatessucks Sep 14 '12 at 7:21
If I use that it doesn't make any plot at all. I just get PLot[\[Piecewise] { {0, Log[1. Sqrt[-1 + 51.3353 En^2]]/Log[10] < 1.7635}, {((4 Log[1. Sqrt[-1 + 51.3353 En^2]])/Log[10]), Log[1. Sqrt[-1 + 51.3353 En^2]]/Log[10] > 1.7635}, {0, \!\(\* TagBox["True", "PiecewiseDefault", AutoDelete->False, DeletionWarning->True]\)} }, {En, 0.001, 1000}] – Thanos Sep 14 '12 at 8:16
Are you really using PLot ? It should be Plot (L not capitalized). – b.gatessucks Sep 14 '12 at 8:26
That is so true... I realised that a couple of hours ago... Thank's for your help! – Thanos Sep 14 '12 at 14:50

2 Answers

up vote 2 down vote accepted

In short, yes it is possible. In your setup \[Beta][En] is complex for En < 0.13957.

I'd do :

f[En_] := Piecewise[{{0, X[En] <= X0}, {4, X[En] > X0}}]

minEn=FindRoot[\[Gamma][En]^2 - 1 == 0, {En, 0.1}][[1,2]]
(* 0.13957 *)

Plot[{X0, X[En], f[En]}, {En, minEn, 10}, PlotRange -> All, 
  PlotStyle -> {Automatic, Automatic, {Red, Thick}}]

plot

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The non real part, is indeed a problem...Actually 0.13957 is a particle's mass. This computation formula doesn't allow me to go bellow that energy, but OK this forum is not about that. Sorry! Thank you very much for your answer! – Thanos Sep 14 '12 at 8:28
up vote 1 down vote

Not really sure if this is what you want to achieve, but anyway

X0 = 1.7635;
X1 = 4.4855;
γ[En_] := En/0.13957
β[En_] := Sqrt[γ[En]^2 - 1]/γ[En]
X[En_] := Log[10, β[En]*γ[En]]
f[En_?NumericQ, h_?NumericQ] := Piecewise[{{0, h < X0}, {4, h > X0}}]
Plot[f[En, X[En]], {En, 1, 10}, Exclusions -> None]

Mathematica graphics

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I can't understand how you relate h with X[En] – Thanos Sep 14 '12 at 8:30
Oh...I just got it! You give f two arguments and when plotting you use X[En] as the second argument! Nice! Thank you very much! – Thanos Sep 14 '12 at 8:32

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