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 > >> > >> > > > > >

**References**:**Re: Free variables in an expression***From:*Bob Hanlon <hanlonr@cox.net>