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Re: Free variables in an expression
*To*: mathgroup at smc.vnet.net
*Subject*: [mg54860] Re: Free variables in an expression
*From*: Sascha Kratky <notvalid at notvalid.com>
*Date*: Fri, 4 Mar 2005 05:07:56 -0500 (EST)
*References*: <200503020626.BAA08074@smc.vnet.net> <d06160$krn$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Andrzej,
with "free" and "bound" I was actually referring to the meaning of
these terms in logic. So the solution you gave is exactly was I was
looking for.
Thanks,
Sascha
In article <d06160$krn$1 at smc.vnet.net>, Andrzej Kozlowski
<akoz at mimuw.edu.pl> wrote:
> 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: [mg54860] 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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