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MathGroup Archive 2005

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Re: Simplfying inside Sqrt

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55120] Re: [mg55106] Simplfying inside Sqrt
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 13 Mar 2005 04:57:38 -0500 (EST)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

You have to assume x to be nonnegative

expr=Sqrt[x^2+x^4];

Simplify[expr,x>=0,
  ComplexityFunction->(Count[{#1},_^_,Infinity]&)]

x*Sqrt[x^2 + 1]

However, FullSimplify goes too far with this ComplexityFunction

FullSimplify[expr,x>=0,
  ComplexityFunction->(Count[{#1},_^_,Infinity]&)]

x*Sqrt[(x - I)*(x + I)]

To prevent FullSimplify from factoring over Gaussian integers use a different 
ComplexityFunction

FullSimplify[expr, x >= 0,
  ComplexityFunction -> (Count[{#1}, _^_, Infinity] +
          Count[{#1}, Complex[__], Infinity] &)]

x*Sqrt[x^2 + 1]


Bob Hanlon

> 
> From: billkavanagh at gmail.com
To: mathgroup at smc.vnet.net
> Date: 2005/03/12 Sat AM 02:36:58 EST
> To: mathgroup at smc.vnet.net
> Subject: [mg55120] [mg55106] Simplfying inside Sqrt
> 
> Hi
> 
> I'm wondering how to tell mathematica that I want terms like
> Sqrt[x^2+x^4] to be x*Sqrt[1+x^2].   I have an expression with a few
> terms like this in it so manually inserting a
> PowerExpand[Sqrt[Expand[x^2+x^4]] is no good to me.
> 
> I've tried a general PowerExpand and Simplify with a Im[x]==0 around
> the whole expression but with no luck.
> 
> Does anybody know how to do this?
> 
> Thanks,
> Bill
> 
> --
> William R. Kavanagh
> http://www.physics.mun.ca/~wkavanag
> 
> 


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