Re: Replacement Rule with Sqrt in denominator

*To*: mathgroup at smc.vnet.net*Subject*: [mg113986] Re: Replacement Rule with Sqrt in denominator*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sat, 20 Nov 2010 06:10:54 -0500 (EST)

On 11/19/10 at 5:11 AM, tmatsoukas at me.com (Themis Matsoukas) wrote: >This replacement rule works, >Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] /. Sqrt[-4 \[Zeta]^2 + (1 >+ \[Rho]^2)^2] -> G >but this doesn't: >1/Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2] /. Sqrt[-4 \[Zeta]^2 + >(1 + \[Rho]^2)^2] -> G >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. It is essential to know Mathematica only does a replacement where there is a literal match when applying replacement rules and the match *must* to the form Mathematica uses to represent the expression. Using FullForm In[10]:= FullForm[1/Sqrt[-4 \[Zeta]^2 + (1 + \[Rho]^2)^2]] Out[10]//FullForm= Power[Plus[Times[-4,Power[\[Zeta],2]],Power[Plus[1,Power[\[Rho],2]],2]],Rational[-1,2]] you can see there is no Sqrt in the denominator. So, your pattern fails to match and no replacement is done.