Re: Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- To: mathgroup at smc.vnet.net
- Subject: [mg63268] Re: [mg63258] Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 22 Dec 2005 00:04:30 -0500 (EST)
- References: <200512210435.XAA14733@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 21 Dec 2005, at 13:35, Steven T. Hatton wrote:
> Is there a way to convince Mathematica to multiply Sqrt[a+b]Sqrt[a-
> b] to
> produce Sqrt[a^2+b^2]?
I hope that it will never become convinced of that!
Perhaps you meant to convince it that Sqrt[a+b]Sqrt[a-b] is Sqrt[a^2
- b^2]. Even so, I hope it will not be convinced *too easily* since:
Sqrt[a - b]*Sqrt[a + b] /. {a -> -1, b -> 0}
-1
Sqrt[(a - b)*(a + b)] /. {a -> -1, b -> 0}
1
However, if you give Mathematica reasonable assumptions then it will
(usually) also be quite reasonable, e.g.
Simplify[Sqrt[a + b]*Sqrt[a - b], {a >= b}]
Sqrt[a^2 - b^2]
Andrzej Kozlowski
- References:
- Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]
- From: "Steven T. Hatton" <hattons@globalsymmetry.com>
- Convincing Mathematica that Sqrt[a+b]Sqrt[a-b]==Sqrt[a^2+b^2]