       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

```

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