Re: How do I create a parametric expression?
- To: mathgroup at smc.vnet.net
- Subject: [mg68579] Re: How do I create a parametric expression?
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Wed, 9 Aug 2006 23:57:42 -0400 (EDT)
- References: <200608090819.EAA21141@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
axlq 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 This is probably best done with a form of algebraic replacement. I seem to revisit this from time to time, for example see the URLs below. But each time the code gets a bit longer. http://forums.wolfram.com/mathgroup/archive/2005/Apr/msg00273.html http://forums.wolfram.com/mathgroup/archive/2002/Jan/msg00354.html The code in those threads will not go inside transcendental functions. So below is a modification that will. replacementFunction[expr_,rep_,vars_] := With[ {num=Numerator[expr],den=Denominator[expr],hed=Head[expr]}, If [PolynomialQ[num,vars] && PolynomialQ[den,vars], PolynomialReduce[num, rep, vars][[2]] / PolynomialReduce[den, rep, vars][[2]] , (* else *) If [Head[hed]===Symbol&&MemberQ[Attributes[hed],NumericFunction], Map[replacementFunction[#,rep,vars]&, expr] , (* else *)expr] ] ] Your example: 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)); It appears to work best here if we do not encapsulate the thing we replace in a square root. In[20]:= InputForm[replacementFunction[expr, a^2+(q-z)^2-R^2, {a,q,z}]] Out[20]//InputForm= -((1 + 2*n)*((-R^2 + k^2*R^4 + q^2*(3 - k^2*R^2) + q*(-6 + 2*k^2*R^2)*z + (3 - k^2*R^2)*z^2)*Cos[k*Sqrt[R^2]] - k*Sqrt[R^2]*(-3*q^2 + R^2 + 6*q*z - 3*z^2)*Sin[k*Sqrt[R^2]])* Sin[((Pi + 2*n*Pi)*z)/L])/(8*Pi*(R^2)^(5/2)*w) Daniel Lichtblau Wolfram Research
- References:
- How do I create a parametric expression?
- From: axlq@spamcop.net (axlq)
- How do I create a parametric expression?