RE: Sqrt problem - restated
- To: mathgroup@smc.vnet.net
- Subject: [mg12188] RE: [mg12164] Sqrt problem - restated
- From: R Finley <trfin@umsmed.edu>
- Date: Fri, 1 May 1998 03:08:40 -0400
Paul,
Your equation still looks funny because of the formatting... I presume
you mean
eq ñ + Sqrt[a-b]*Sqrt[a+b]/Sqrt[a^2-b^2]
One way to do it is to use the rule (I believe Bob Hanlon wrote this in
his response to your first note)
subst ÿqrt[x_]*Sqrt[y_] -> Sqrt[x*y] Simplify[eq/.subst]
And you will get your answer of 2. Of course, it is up to you to ensure
that the x and y actually DO satisfy the qualifications that you give
them (real, positive). If there are complicated expressions under
the square root, this might not be obvious.
Hope that helps.
RF
-----Original Message-----
From: PAUL REISER [SMTP:REISER@HRTS.NRL.NAVY.MIL] To:
mathgroup@smc.vnet.net
Sent: Monday, April 27, 1998 12:47 AM To: mathgroup@smc.vnet.net
Subject: [mg12164] Sqrt problem - restated
That square root problem was a mis-statement on my part, but thank you
to everyone who helped. The problem I am having is something like how
do I get "2" out of this:
Sqrt[a - b] Sqrt[a + b]
1 + -----------------------
2 2
Sqrt[a - b ]
I'm starting to understand why Mathematica doesn't deal with this
because of the multiple value of square root, but if I know everything
under the square roots are positive, can I reduce this to 2 somehow
without substitution? (substitution would be prohibitively tedious in
my case)
Thanks for any help
Paul Reiser