[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: Replacement Rule with Sqrt in denominator**
Next by Date:
**Re: How to work with In?**
Previous by thread:
**Re: Replacement Rule with Sqrt in denominator**
Next by thread:
**Re: Replacement Rule with Sqrt in denominator**
| |