Re: Re: Big problem in solving radicals.
- To: mathgroup at smc.vnet.net
- Subject: [mg41789] Re: [mg41761] Re: Big problem in solving radicals.
- From: Selwyn Hollis <selwynh at earthlink.net>
- Date: Thu, 5 Jun 2003 07:31:31 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Perhaps confusion arises from the term "general solution." A better term is "formal solution." When Solve[x^(1/2) + a == 0, x] gives {{x -> a^2}}, it's saying that *if* there's a solution, this is it. It's not saying that the solution is valid for all possible a. ----- Selwyn Hollis http://www.math.armstrong.edu/faculty/hollis On Wednesday, June 4, 2003, at 08:34 AM, Jens-Peer Kuska wrote: > Hi, > > > {{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. > > Regards > Jens > > 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 > >