Re: Big problem in solving radicals.
- To: mathgroup at smc.vnet.net
- Subject: [mg41769] Re: [mg41744] Big problem in solving radicals.
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Wed, 4 Jun 2003 08:34:52 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I am not sure what you mean by "the complex solution" in this case.
There is no "solution" that is valid universally for all values of a.
Moreover, what you seem to be asking for, that is a method for solving
such equations with arbitrary parameters in complicated cases is
impossible now and almost certainly will never be possible. An radical
expression with parameters, say ((a+(1+b)^(1/3))^(1/5) is basically a
multi-branched function in the complex plane. When a an b are numeric
one can use the technique of root isolation and root ordering to give
this a unique meaning, e.g.
In[9]:=
RootReduce[(a + (1 + b)^(1/3))^(1/5) /. {a -> 2, b -> 3}]
Out[9]=
Root[-12 + 12*#1^5 - 6*#1^10 + #1^15 & , 1]
is a single well defined complex number. But there is no such
description for the expression with parameters. That means that, except
in very special cases, there is no way to hope to be able to write out
any "solution" to the sort of equation you are talking about. And any
solutions that one can arrive is special cases will only be valid for
parameters lying in certain parts of the complex plane.
Nevertheless, I do think you have come close to spotting a bug. I do
not thing the answer Mathematica gives in your case is a bug, but the
following seems to me to be one:
In[10]:=
Solve[x^(1/2) + a == 0, x, VerifySolutions -> True]
Out[10]=
{{x -> a^2}}
I though VerifySolutions was supposed to rule out "solutions" of this
kind?
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/
On Tuesday, June 3, 2003, at 08:13 pm, 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
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> E-Mail: davide @ astromeccanica.it
> E-Mail: del vento @ ifsi . rm . cnr . it
>
>
>