Re: Pattern Matching: Keep those summands not depending on x
- To: mathgroup at smc.vnet.net
- Subject: [mg63828] Re: Pattern Matching: Keep those summands not depending on x
- From: "ben" <benjamin.friedrich at gmail.com>
- Date: Tue, 17 Jan 2006 04:33:24 -0500 (EST)
- References: <dq7us0$40n$1@smc.vnet.net><dqacpv$1dd$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Thank you all for your solutions
Valeri Astanoff wrote me the shortest one:
Select[exp0, FreeQ[#, x] &].
However, the suggestion
Coefficient[exp0,x,0]
does not work, just think of exp0=1+Sin[x].
Peter Pein schrieb:
> ben schrieb:
> > Dear all,
> >
> > I have a complicated sum and want to extract all summands not depending
> > on x.
> >
> > I have a clumsy way to do that, but I wonder whether there is a better
> > solution.
> >
> > So that is what I have done:
> >
> > (* The sum *)
> >
> > exp0 = 1+x;
> >
> > (* Thread a dummy function foo1 over the summands using FixedPoint *)
> > exp1=foo1[exp0];
> >
> > foo2[exp_]:=exp/.{foo1[a_+b_]\[Rule]foo1[a]+foo1[b]}
> >
> > exp2=FixedPoint[foo2/.{foo1[a_+b_]\[Rule]foo1[a]+foo1[b]},
> > exp1]/.{foo1[exp_]\[Rule]foo3[exp]};
> >
> > (* Replace dummy function foo1 by another one f003
> > that keeps the terms not depending on
> > x and drops the others *)
> > foo3[exp_]:=If[FreeQ[exp,s],exp,0]
> >
> > (* The result *)
> > exp2
> >
> > Any help is appreciated
> >
> > Ben
> >
>
> Hi Ben,
>
> you don't need dummy functions. Let's take a slightly more complicated expression:
>
> (1 + x + y)*(z + x) /. a_Plus :> Plus @@ Select[List @@ a, FreeQ[#1, x] & ]
>
> --> (1 + y)*z
>
> or if your problem allows this:
>
> (1 + x + y)*(z + x) /. x -> 0
>
> --> (1 + y)*z
>
>
> Peter