Re: Re: Question about "Reduce"
- To: mathgroup at smc.vnet.net
- Subject: [mg61936] Re: [mg61921] Re: Question about "Reduce"
- From: Chris Chiasson <chris.chiasson at gmail.com>
- Date: Sat, 5 Nov 2005 01:52:19 -0500 (EST)
- References: <dkco86$qcp$1@smc.vnet.net> <200511041011.FAA14948@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Sometimes I think I notice (in the output of Reduce) conditions that
should evaluate to true given the assumptions I supplied. So, I am not
suprized by your solution.
On 11/4/05, Stanley Rabinowitz <stan.rabinowitz at comcast.net> wrote:
> I changed my "Reduce" command to
>
> Reduce[ForAll[{u, v}, 0 <= u <= Sqrt[3] && 0 <= v <= Sqrt[3] && 0 <= 1
> - u v<= (u + v)Sqrt[3], ineq]] // FullSimplify
>
> and this time I got the correct result.
> So I am thinking perhaps this is a bug in Mathematica.
> Since u>=0 and v>=0, the conditions
> 0 <= (1 - u v)/(u + v)<= Sqrt[3] and 0 <= 1 - u v<= (u + v)Sqrt[3]
> should be equivalent. However, the latter condition refers only to
> polynomials,
> so may be easier for Mathematica.
>
>
--
http://chrischiasson.com/contact/chris_chiasson
- References:
- Re: Question about "Reduce"
- From: "Stanley Rabinowitz" <stan.rabinowitz@comcast.net>
- Re: Question about "Reduce"