RE: Simplify using custom function

*To*: mathgroup at smc.vnet.net*Subject*: [mg47697] RE: [mg47676] Simplify using custom function*From*: "DrBob" <drbob at bigfoot.com>*Date*: Thu, 22 Apr 2004 02:38:49 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

expr = -1 + 2*x + 2*E^x; f[x_] = x + E^x Off[Eliminate::ifun] Eliminate[{expr == y, f[x] == z}, x] Block[{f}, result = First[%] /. z -> f[x]]; resultE^x + x -1 + 2*z == y -1 + 2*(E^x + x) ?result Global`result result = -1 + 2*f[x] DrBob www.eclecticdreams.net -----Original Message----- From: Jens [mailto:jens at tnntw10.tn.tudelft.nl] To: mathgroup at smc.vnet.net Subject: [mg47697] [mg47676] Simplify using custom function What I'm trying to achieve is to find a certain function in a more complicated expression: example: complicated expression: -1+2 x +2 e^x function: f[x_]=x+e^x Mathematica should return: -1+2 f[x] I've been looking at TransformationFunctions and FullSimplify, but did not succeed in defining a transformation function so far. How do I do that? (It's Mathematica 5.0) Jens