RE: How do I create a parametric expression?

*To*: mathgroup at smc.vnet.net*Subject*: [mg68575] RE: How do I create a parametric expression?*From*: "David Park" <djmp at earthlink.net>*Date*: Wed, 9 Aug 2006 23:57:35 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Alex, You could do something like the following. expr = -((1 + 2*n)*((a^4*k^2 + a^2*(-1 + k^2*(q - z)^2) + 2*(q - z)^2)*Cos[k*Sqrt[a^2 + (q - z)^2]] - k*(a^2 - 2*(q - z)^2)*Sqrt[a^2 + (q - z)^2]* Sin[k*Sqrt[a^2 + (q - z)^2]])* Sin[((1 + 2*n)*Pi*z)/L])/ (8*Pi*w*(a^2 + (q - z)^2)^(5/2)); expr /. (q - z)^2 -> r^2 - a^2 FullSimplify[%, r > 0] % /. r -> Sqrt[a^2 + (q - z)^2] Simplify[%, a < 0] I don't know if you can assume that r > 0. You will probably get a better answer from the algebraists in the group. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: axlq [mailto:axlq at spamcop.net] To: mathgroup at smc.vnet.net I'm trying to figure out how to simplify a large expression so that it's expressed in terms of a sub-expression that's factored into the larger one. My expression looks like this: -((1 + 2*n)*((a^4*k^2 + a^2*(-1 + k^2*(q - z)^2) + 2*(q - z)^2) *Cos[k*Sqrt[a^2 + (q - z)^2]] - k*(a^2 - 2*(q - z)^2) *Sqrt[a^2 + (q - z)^2]*Sin[k*Sqrt[a^2 + (q - z)^2]]) *Sin[((1 + 2*n)*Pi*z)/L])/(8*Pi*w*(a^2 + (q - z)^2)^(5/2)) Now, I *know* there are places in there were Sqrt[a^2+(q-z)^2] occurs, either by itself or raised to various powers. If I want to define R:=Sqrt[a^2+(q-z)^2] ...then how can I make Mathematica re-state my expression in terms of R? The ReplaceRepated[] function doesn't seem to do the job. I need to do this because I am translating the expressions into Visual Basic code for an Excel application, and it would be nice to find groupings of terms repeated throughout the expression that I need to calculate only once. -Alex