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
>
>