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MathGroup Archive 2002

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Re: How to ...?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36334] Re: [mg36309] How to ...?
  • From: BobHanlon at aol.com
  • Date: Mon, 2 Sep 2002 04:08:42 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 8/31/02 1:58:36 AM, berlusconi_pagliusi at fis.unical.it
writes:


> I'd like to use Mathematica 4.0 to write a function having different
> expressions in different domain's intervals.
> Let's say:
>
> F[x_]=  x^2 if 0<x<6
>        x+1 if x>=6
>
> I know It's a stupid syntax problem, but I really do not know how/where to
> search the solution on the Mathematica Book
>

Just for grins here are several methods:

f1[x_/;x<=0] := 0 ;

f1[x_/;0<x<6] := x^2 ;

f1[x_/;x>=6] := x+1;


f2[x_] :=

     x^2*UnitStep[x]+(x+1-x^2)*UnitStep[x-6];


f3[x_] := Which[

       x<=0, 0,

       0<x<6, x^2,

       x>=6, x+1];


f4[x_] := If[0<x<6,x^2,

       If[x>=6, x+1,0]];


f5[x_] := Switch[x,

     _?(#<=0&), 0,

     _?(0<#<6&), x^2,

     _?(#>=6&), x+1 ]


f6[x_?NumericQ] :=

     {0,x^2,x+1}[[Position[

             {x<=0,0<x<6,x>=6}, True][[1,1]]]];


Needs["Calculus`Integration`"];

(* needed for definition of Boole *)


f7a[x_?NumericQ] := Evaluate[

       (Boole /@ {0<x<6,x>=6}).

         {x^2,x+1}];


Off[Part::pspec];

f7b[x_?NumericQ] := Evaluate[

       {0,x^2,x+1}[[1+Tr[Boole /@ {x>0, x>=6}]]]];


f8[x_?NumericQ] := Cases[

         {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}},

         {True, z_} :>z][[1]];


f9[x_?NumericQ] := DeleteCases[

         {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}},

         {False, z_}][[1,2]];


f10[x_?NumericQ] := Select[

         {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}},

         First[#]&][[1,2]];


f11[x_?NumericQ] := Last[Sort[

           {{x<=0, 0}, {0<x<6, x^2}, {x>=6, x+1}}]][[2]];


f12[x_?NumericQ] := Module[{n=1},

       While[{x<6,x<0, False}[[n]], n++];

       {x+1,x^2,0}[[n]]];


Generating some test points:

ts = {Random[Real,{-5,0}],0, Random[Real,{0,6}],6,Random[Real,{6,15}]};


Checking the different representations

Equal[(# /@ pts)& /@ {f1,f2,f3,f4,f5,f6,f7a,f7b,f8,f9,f10,f11,f12}]


True

To pick a favorite, look at how the different definitions behave.


Of the definitions that evaluate with symbolic input, only f2 and f4
simplify with assumptions
.   For example,

FullSimplify[#[x]& /@ {f2,f4}, 1<x<3]


f2 through f5 respond immediately to differentiation


#'[x]& /@ {f2,f3,f4,f5} // Simplify //ColumnForm


Only f2 responds immediately to integration


Integrate[f2[x],x]//Simplify

Consequently, f2 (UnitStep) appears to be the most versatile.


Bob Hanlon
Chantilly, VA   USA



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