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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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