Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63286] Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
  • From: "Jean.Pellegri" <Jean.Pellegri at wanadoo.fr>
  • Date: Fri, 23 Dec 2005 05:08:14 -0500 (EST)
  • References: <dod6vb$4fi$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Bonjour

You can define a rule :

In[1 := rule=Sqrt[a_- b_]Sqrt[a_+ b_]->Sqrt[a^2 - b^2]


In[2] := Sqrt[a-b]Sqrt[a+b] /. rule

Out[2] := Sqrt[a^2-b^2]



Jean


"Steven T. Hatton" <hattons at globalsymmetry.com> a écrit dans le message de 
news: dod6vb$4fi$1 at smc.vnet.net...
> Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a-b] to
> produce Sqrt[a^2+b^2]?
> -- 
> The Mathematica Wiki: http://www.mathematica-users.org/
> Math for Comp Sci http://www.ifi.unizh.ch/math/bmwcs/master.html
> Math for the WWW: http://www.w3.org/Math/
> 



  • Prev by Date: Re: Returning an empty sequence
  • Next by Date: Re: Returning an empty sequence
  • Previous by thread: Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
  • Next by thread: "Alternating" function