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RE: How do I create a parametric expression?


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

From: axlq [mailto:axlq at]
To: mathgroup at

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

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


...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.


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