[Date Index]
[Thread Index]
[Author Index]
Re: Convincing Mathematica that Sqrt[a+b]Sqrt[ab]==Sqrt[a^2+b^2]
 To: mathgroup at smc.vnet.net
 Subject: [mg63272] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[ab]==Sqrt[a^2+b^2]
 From: Pratik Desai <pdesai1 at umbc.edu>
 Date: Thu, 22 Dec 2005 00:04:33 0500 (EST)
 References: <200512210435.XAA14733@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
Steven T. Hatton wrote:
>Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[ab] to
>produce Sqrt[a^2+b^2]?
>
>
You mean
Sqrt[a^2b^2] not Sqrt[a^2+b^2]
To show the latter is beyond my capabilities :)
After trying to convince mathematica for about 15 minutes, I decided to
take matters in to my own hands
powexp = Sqrt[x__]*Sqrt[y__] > Sqrt[x*y]
Sqrt[ab]*Sqrt[a+b]/.powexp//Simplify
>>Sqrt[a^2b^2]
This is ofcourse assuming that a and b are real, which is why
Mathematica is reluctant to carry out the expansion in the first place
Hope this helps
Pratik

Pratik Desai
...Moderation, as well as Regularity of Thinking, so much to be wished for in the Heads of those who imagine they come into the World only to watch and govern it?s Motion
Gulliver's Travels
by Jonathan Swift
Prev by Date:
Re: Convincing Mathematica that Sqrt[a+b]Sqrt[ab]==Sqrt[a^2b^2]
Next by Date:
Re: "Alternating" function
Previous by thread:
Re: Convincing Mathematica that Sqrt[a+b]Sqrt[ab]==Sqrt[a^2b^2]
Next by thread:
Re: Convincing Mathematica that Sqrt[a+b]Sqrt[ab]==Sqrt[a^2+b^2]
