Re: Replacement Rule with Sqrt in denominator

*To*: mathgroup at smc.vnet.net*Subject*: [mg114003] Re: Replacement Rule with Sqrt in denominator*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sat, 20 Nov 2010 06:14:12 -0500 (EST)

Replacements are done on the FullForm. Look at the FullForm to understand the differences. expr1 = Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2]; expr2 = 1/Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2]; rule1 = Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] -> G; expr1 // FullForm rule1 // FullForm expr2 // FullForm Note that in the FullForm for both expr1 and rule 1 that the pattern is Power[_, Rational[1, 2]] so the replacement is made. However, in the FullForm for expr2 the pattern is Power[_,Rational[-1, 2]] so there is no match. You can generalize the rule rule2 = (-4 \[Zeta]^2 + (1 + \[Rho]^2)^2)^Rational[a_, 2] -> G^a; {expr1, expr2} /. rule2 {G, 1/G} Or avoid the Power rule3 = (-4 \[Zeta]^2 + (1 + \[Rho]^2)^2) -> G^2; {expr1, expr2} /. rule3 // Simplify[#, G > 0] & {G, 1/G} Bob Hanlon ---- Themis Matsoukas <tmatsoukas at me.com> 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