Re: Patterns with conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg116896] Re: Patterns with conditions
• From: Leonid Shifrin <lshifr at gmail.com>
• Date: Fri, 4 Mar 2011 03:36:39 -0500 (EST)

```Hi Jakub,

As you noted, the error message is because the first pattern doe not match
on the <tab>, and the second
definition applies, but its r.h.s. does not guard against the x = 0 case.
All you have to do is to add

SetAttributes[sinc, Listable];

(although this is not necessary when
you add Listable)  restrict the pattern of your second rule to numeric
arguments, like x_?NumericQ.

Regards,
Leonid

2011/3/3 =C5 er=C3=BDch Jakub <Serych at panska.cz>

> 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]
>
>

```

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