Re: How do I create a parametric expression?

*To*: mathgroup at smc.vnet.net*Subject*: [mg68562] Re: How do I create a parametric expression?*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Wed, 9 Aug 2006 23:56:59 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

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)); expr2=Simplify[expr/.a^2+(q-z)^2:>R^2,R>0] -((1/(8*Pi*R^5*w))*((2*n + 1)* ((k^2*a^4 + (k^2*(q - z)^2 - 1)*a^2 + 2*(q - z)^2)* Cos[k*R] - k*R*(a^2 - 2*(q - z)^2)*Sin[k*R])* Sin[((2*n + 1)*Pi*z)/L])) expr3=Simplify[expr/.(q-z)^2:>R^2-a^2,R>0] -((1/(8*Pi*R^5*w))*((2*n + 1)* (((k^2*R^2 - 3)*a^2 + 2*R^2)*Cos[k*R] + k*R*(2*R^2 - 3*a^2)*Sin[k*R])* Sin[((2*n + 1)*Pi*z)/L])) Bob Hanlon ---- axlq <axlq at spamcop.net> wrote: > > 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 >