Re: Automatic generation of piecewise functions
- To: mathgroup@smc.vnet.net
- Subject: [mg11603] Re: Automatic generation of piecewise functions
- From: Allan Hayes <hay@haystack.demon.co.uk>
- Date: Tue, 17 Mar 1998 10:43:37 -0500
- References: <6ek6o3$jen@smc.vnet.net>
William F. Campbell wrote:
>
> I am trying to generate a single function f[x], which has different
> definitions depending on the values of x (i.e. a piecewise function).
> I proceeded as follows:
>
> Clear[f];
> sampleÚble[f[x_]«i]*x+b[i]/;Evaluate[x>-1&&x<-,{i,2}]
>
> The output returned looked correct:
> {b[1]+a[1] x/;x>&x< b[2]+a[2] x/;x>&x<-
>
> When I looked at the definition for f, it seemed correct as well: ?f
>
> Global `f
>
> f[x_]»1]+a[1] x/;x>&x<
>
> f[x_]»2]+a[2] x/;x>&x<-
>
> But when I attempted to use the definition, f[0.5]
>
> 0.5 a[1]+b[1]/;0.5>&0.5<
>
> whose full form is
> FullForm[f[0.5]]
>
> Condition[Plus[Times[0.5,a[1]],b[1]],
> And[GreaterEqual[0.5,0],LessEqual[0.5,1]]]
>
> Why doesn't the Condition evaluate? How can I change the construct used
> in sample above to obtain a working definition for a piecewise
> function? BTW, If I fail to put Evaluate[] around the condition in the
> definition of sample above, the output is even further from what I
> want, {b[1]+a[1] x/;x>-1&&x<,b[2]+a[2] x/;x>-1&&x<-. --
> Bill Campbell Correlation is not cause.
Bill:
Some points.
1) With Set ( Condition must be on left side:
r[x_] x/;x>
l[x_]/;x> x;
{r[.5],l[.5]}
{1.5/;0.5*0,1.5}
With SetDelayed (:Condition can be on either side:
rd[x_] :x/;x>
ld[x_]/;x> :x;
{rd[.5],ld[.5]}
{1.5,1.5}
2) Then there is the matter of evaluating the condition to get the
values of the index i inside it - this you solve by using Evaluate.
Here I combine this with moving the condition to the left AND inside f
ClearAll[f];
Table[f[x_/;Evaluate[i-1<x<-]
[i]*x,{i,2} ];
{f[.5],f[1.5]}
{0.5 a[1],1.5 a[2]}
If f has two places the condition may have to be outside f and then an
extra
Evaluate is needed
ClearAll[f];
Table[Evaluate[f[x_,y_]/;Evaluate[i-1<x+y<-]
[ i]*x,{i,2} ];
{f[.5,0],f[1.5,0]}
{0.5 a[1],1.5 a[2]}
However, forcing evaluation may do more than we want; and using
replacement or
With may suit us better (though the former may replace too much and the
latter
may get tangled up with renaming in scoping constructs):
ClearAll[f];
Table[
Unevaluated[
f[x_,y_]/;i-1<x+y<-a[ i]*x
]/.i->j,{j,2}
]; (*Unevaluated puts assignment after replacement*)
{f[.5,0],f[1.5,0]}
{0.5 a[1],1.5 a[2]}
ClearAll[f];
Table[
With[{i-,
f[x_,y_]/;i-1<x+y<-a[ i]*x
],
{i,2}
];
{f[.5,0],f[1.5,0]}
{0.5 a[1],1.5 a[2]}
--
Allan Hayes
Training and Consulting
Leicester, UK
hay@haystack.demon.co.uk
http://www.haystack.demon.co.uk
voice: +44 (0)116 271 4198
fax: +44 (0)116 271 8642