Re: Big problem in solving radicals.
- To: mathgroup at smc.vnet.net
- Subject: [mg41794] Re: Big problem in solving radicals.
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Thu, 5 Jun 2003 07:31:36 -0400 (EDT)
- Organization: The University of Western Australia
- References: <bbi16p$7c1$1@smc.vnet.net> <bbkq1p$hi5$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <bbkq1p$hi5$1 at smc.vnet.net>,
Jens-Peer Kuska <kuska at informatik.uni-leipzig.de> wrote:
> {{x -> a^2}}
>
> *is* the general solution, nobody say that x (or a)
> must be real.
>
> There is no way to ask Mathematica for only a real
> solution in symbolic expressions.
What about
Experimental`CylindricalAlgebraicDecomposition[x^(1/2) + a == 0, {a,x}]
a <= 0 && x == a^2
Cheers,
Paul
> Davide Del Vento wrote:
> >
> > Consider the following equation
> >
> > 1/2
> > x + a = 0
> >
> > If you try to solve it with "Solve" you get
> >
> > 2
> > x = a
> >
> > Of course, you know, this is not a general solution, e.g. if a>0 there
> > isn't any (real) solution, and the complex solution is NOT the one
> > printed by Mathematica.
> >
> > In the case of this example the problem is obvious and one can track
> > it by hand, but what's about bigger equations with many solutions?
> > Mathematica claims that "Solve" makes special assumptions about the
> > parameters in the equation, so I was ready to such behaviour. I tested
> > "Reduce"
> > that should solve equation, giving explicitely the range of the
> > parameters where the solutions are defined. Unfortunately it doesn't
> > work right too.
> >
> > ;Davide Del Vento
> >
> > CNR Istituto Fisica Spazio Interplanetario
> > via del Fosso del Cavaliere, 100 / IT-00133 / Rome
> > Phone: +390649934357
> > Fax: +390649934383
> > Mobile: +393288329015
> > E-Mail: davide @ astromeccanica.it
> > E-Mail: del vento @ ifsi . rm . cnr . it
>
--
Paul Abbott Phone: +61 8 9380 2734
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