Re: Patterns with conditions
- To: mathgroup at smc.vnet.net
- Subject: [mg116929] Re: Patterns with conditions
- From: Peter Pein <petsie at dordos.net>
- Date: Fri, 4 Mar 2011 03:42:37 -0500 (EST)
- References: <iknsln$kcn$1@smc.vnet.net>
Am 03.03.2011 12:05, schrieb ?erých Jakub:
> 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]
>
Hi,
if you want to get the behaviour of Sinc, you should drop the \[Pi]s in
the definition. In Version 8 works:
In[1]:= sinc[x_] := 1 /; x == 0;
sinc[x_] := Sin[x]/x;
SetAttributes[sinc, Listable];
In[4]:= sinc[Range[0, 2, 1/3]*Pi]
Out[4]= {1, (3*Sqrt[3])/(2*Pi), (3*Sqrt[3])/(4*Pi), 0,
-((3*Sqrt[3])/(8*Pi)), -((3*Sqrt[3])/(10*Pi)), 0}
And you're right: any list is != 0. That is the reason for the attribute
Listable.
hth,
Peter