RE: Automatic generation of piecewise functions
- To: mathgroup@smc.vnet.net
- Subject: [mg11570] RE: [mg11549] Automatic generation of piecewise functions
- From: jmt <jmthomas@cybercable.tm.fr>
- Date: Tue, 17 Mar 1998 10:42:54 -0500
- Organization: TeA
Maybe I did not understand your question but why don't you define your
function this way: f[x_]=x a[Floor@x]+b[Floor@x]
Hope this helps
----------------------------------------------- Jean-Marie THOMAS
Conseil et Audit en Ingenierie de Calcul www.cybercable.tm.fr/~jmthomas
------------------------------------------------
-----Message d'origine-----
De: William F. Campbell [SMTP:valentin@wam.umd.edu] Date: samedi 14 mars
1998 19:56
A: mathgroup@smc.vnet.net
Objet: [mg11549] Automatic generation of piecewise functions
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];
sample=Table[f[x_]=a[i]*x+b[i]/;Evaluate[x>=i-1&&x<=i],{i,2}]
The output returned looked correct:
{b[1]+a[1] x/;x>=0&&x<=1,b[2]+a[2] x/;x>=1&&x<=2}
When I looked at the definition for f, it seemed correct as well: ?f
Global `f
f[x_]=b[1]+a[1] x/;x>=0&&x<=1
f[x_]=b[2]+a[2] x/;x>=1&&x<=2
But when I attempted to use the definition, f[0.5]
0.5 a[1]+b[1]/;0.5>=0&&0.5<=1
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>=i-1&&x<=i,b[2]+a[2] x/;x>=i-1&&x<=i}. -- Bill
Campbell Correlation is not cause.