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MathGroup Archive 2005

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Re: Re: Free variables in an expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54828] Re: [mg54794] Re: [mg54758] Free variables in an expression
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 2 Mar 2005 22:29:17 -0500 (EST)
  • References: <200503020626.BAA08074@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

The problem is that the OP completely failed to explain the meaning of  
"free" and "bound". With the usual meaning of these terms in  logic the  
following sentence:

expr=Exists[y, Element[y, Reals], x*y - 1 == 0]

contains just one free variable x.  In this case however


Cases[expr, _Symbol?( !NumericQ[#1] & ), Infinity]

{y, y, Reals, y, x}

which is of course wrong. However,


Union[Cases[Resolve[expr],_Symbol? 
(FreeQ[Attributes[#],Protected]&),Infinity]]

{x}

returns the correct answer.

I have to admit I rather doubt that this is what the OP meant, but if  
not than his use of "free" and "bound" is non-standard.

Andrzej Kozlowski


On 2 Mar 2005, at 07:26, Bob Hanlon wrote:

> expr=Log[6x]+y/3-Exp[z]*Pi-3*I;
>
> Cases[expr, _Symbol?(!NumericQ[#]&),Infinity]
>
> {z,y,x}
>
>
> Bob Hanlon
>
>>
>> From: Sascha Kratky <notvalid at notvalid.com>
To: mathgroup at smc.vnet.net
>> Date: 2005/03/01 Tue AM 01:58:13 EST
>> Subject: [mg54828] [mg54794] [mg54758] Free variables in an expression
>>
>> Is there a Mathematica built-in function that gives a list of free
>> (unbound) variables in an arbitrary expression?
>>
>> Thanks,
>> Sascha
>>
>>
>
>


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