Re: Patterns with conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg116903] Re: Patterns with conditions
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Fri, 4 Mar 2011 03:37:55 -0500 (EST)

```Here are a couple of improvements:

ClearAll[sinc]
SetAttributes[sinc, Listable]
sinc[0 | 0.] = 1;
sinc[x_] := Sin[\[Pi] x]/(\[Pi] x);
tab = {0,
0., \[Pi]/3, \[Pi]/2, \[Pi] 2/3, \[Pi], \[Pi] 4/3, \[Pi] 5/3,
2 \[Pi]};
sinc@tab

or

ClearAll[sinc]
sinc[0 | 0.] = 1;
sinc[x_List] := sinc /@ x
sinc[x_] := Sin[\[Pi] x]/(\[Pi] x);
tab = {0,
0., \[Pi]/3, \[Pi]/2, \[Pi] 2/3, \[Pi], \[Pi] 4/3, \[Pi] 5/3,
2 \[Pi]};
sinc@tab

Bobby

On Thu, 03 Mar 2011 04:58:30 -0600, =A6er=FDch 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]
>

--
DrMajorBob at yahoo.com

```

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