Re: How to pick all terms which have are multiplied with a factor, e.g., 2

*To*: mathgroup at smc.vnet.net*Subject*: [mg35476] Re: [mg35463] How to pick all terms which have are multiplied with a factor, e.g., 2*From*: BobHanlon at aol.com*Date*: Sun, 14 Jul 2002 06:19:53 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

My initial response did not handle negative terms and pure numerical constants. This should be better: intFactor[expr_] := Module[{u}, Fold[(Collect[#1 /. {#2*se_. -> u*se, -#2*se_. -> -u*se}, u] /. u -> #2)&, expr, Union[Cases[expr, n_Integer*_ -> n]]]]; intFactor[2 a b^2 c^3+13 x y a b+2 d^-1+2-13z] 2*(1 + a*b^2*c^3 + 1/d) + 13*(a*b*x*y - z) Bob Hanlon In a message dated 7/13/02 7:02:01 AM, writes: >In a message dated 7/13/02 4:26:31 AM, mgi at vt.edu writes: > >>I would like to simplify a large symbolic summation. I know that terms >>which >>are multiplied with a common integer factor can ultimately be combined >>and >>simplified. >> >>As an example: from the sum >> >> 2 a b^2 c^3 + 13 x y a b + 2 d^-1 >> >>I would like to pick 2 a b^2 c^3 + 2 d^-1 to combine them; and similarly >>for >>other common integer factors. I suspect that patterns would help, but >I >>did >>not manage to devise a correct one. > >intFactor[expr_] := Module[{u}, > Fold[(Collect[#1 /. #2*se_ -> u*se, u] /. u->#2)&, expr, > Union[Cases[expr, n_Integer*_ -> n]]]]; > >expr = 2 a b^2 c^3+13 x y a b+2 d^-1; > >{intFactor[expr], Simplify[expr]} > >{2*(a*b^2*c^3 + 1/d) + 13*a*b*x*y, > 2/d + a*b*(2*b*c^3 + 13*x*y)} > >LeafCount /@ % > >{21, 20} > >expr = 2 a b^2 c^3+13 x y a b+2 d^-1+13z; > >{intFactor[expr], Simplify[expr]} > >{2*(a*b^2*c^3 + 1/d) + 13*(a*b*x*y + z), > 2/d + a*b*(2*b*c^3 + 13*x*y) + 13*z} > >LeafCount /@ % > >{24, 23} > >As measured by LeafCount, Simplify works better. >