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MathGroup Archive 2006

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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
> 


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