Re: Patterns with conditions
- To: mathgroup at smc.vnet.net
- Subject: [mg116918] Re: Patterns with conditions
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 4 Mar 2011 03:40:37 -0500 (EST)
tab = {0, 1/3, 1/2, 2/3, 1, 4/3, 5/3, 2 };
ClearAll[sinc]
SetAttributes[sinc, Listable]
sinc[x_ /; x == 0] := 1;
sinc[x_] := Sin[Pi x]/(Pi x);
sinc[tab] == Sinc[Pi tab]
True
ClearAll[sinc]
sinc[x_ /; x == 0] := 1;
sinc[x_] := Sin[Pi x]/(Pi x);
sinc[x_List] := sinc /@ x;
sinc[tab] == Sinc[Pi tab]
True
Bob Hanlon
---- "=C5 er=C3=BDch Jakub" <Serych at panska.cz> wrote:
==========================
Dear Mathematica group,
I'm playing with function definitions and patterns based multiple definition of the function. I have defined this function:
sinc[x_ /; x == 0] := 1;
sinc[x_] := Sin[\[Pi] x]/(\[Pi] x);
(I know, that Mathematica has Sinc function defined, it's just the test.)
It works fine for let's say sinc[\[Pi]], even for sinc[0]. But if I define the table:
tab = {0, \[Pi]/3, \[Pi]/2, \[Pi] 2/3, \[Pi], \[Pi] 4/3, \[Pi] 5/3,
2 \[Pi]};
and I let my function evaluate the results sinc[tab], it returns error messages:
Power::infy: Infinite expression 1/0 encountered. >> and
Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >>
I can understand, that "tab" doesn't fit to pattern condition /; x==0, also I know, that it is possible to Map my function to table
Map[sinc, tab] and it works fine.
I can imagine solution with IF[x==0,1, Sin[\[Pi] x]/(\[Pi] x), but my question is: Is it possible to make my function fully Listable using just pattern conditions?
Thanks for responses
Jakub
P.S. Code in one block for easy copying:
sinc[x_ /; x == 0] := 1;
sinc[x_] := Sin[\[Pi] x]/(\[Pi] x);
tab = {0, \[Pi]/3, \[Pi] 2/3, \[Pi], \[Pi] 4/3, \[Pi] 5/3, 2 \[Pi]};
sinc[\[Pi]]
sinc[0]
sinc[tab]
Map[sinc, tab]