Re: Replacement Rule with Sqrt in denominator

*To*: mathgroup at smc.vnet.net*Subject*: [mg113980] Re: Replacement Rule with Sqrt in denominator*From*: Peter Breitfeld <phbrf at t-online.de>*Date*: Sat, 20 Nov 2010 06:09:47 -0500 (EST)*References*: <ic5igm$44p$1@smc.vnet.net>

Themis Matsoukas wrote: > This replacement rule works, > > Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] /. > Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] -> G > > G > > but this doesn't: > > 1/Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] /. > Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] -> G > > 1/Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] > > The only difference is that the quantity to be replaced is in the denominator. If I remove Sqrt from the replacement rule it will work but I don't understand why a square root in the denominator (but not in the numerator) causes the rule to fail. > > Themis > Have a look at the FullForm, which is always used by ReplaceAll: In your first example the FullForm looks like: ff1=Power[Plus[Times[-4,Power[\[Zeta],2]],Power[ Plus[1,Power[\[Rho],2]],2]],Rational[1,2]] In the second one: ff2=Power[Plus[Times[-4,Power[\[Zeta],2]],Power[ Plus[1,Power[\[Rho],2]],2]],Rational[-1,2]] the little but important difference is the "-" Sign in the last one. In the first example the replacement takes place, because the FullForm matches. You do ff1/.ff1->G But in the second one it doesn't match because you try ff2/.ff1->G You can do: ff2/. (1/ff1)->G (no need to use FullForm here) -- _________________________________________________________________ Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de