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MathGroup Archive 2006

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Re: Why is the negative root?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69537] Re: Why is the negative root?
  • From: p-valko at tamu.edu
  • Date: Fri, 15 Sep 2006 06:44:49 -0400 (EDT)
  • References: <200609130803.EAA18412@smc.vnet.net><eebd6d$l97$1@smc.vnet.net>

"Why do you need to use ToRadicals here?"

I need the positive root(s) of  the equation  z^3 - z^2 - b*z-1==0
where b>0. (This is a cubic equation of state problem.)

I know the answer: there is only one positive root and it is given by

1/3 - (2^(1/3)*(-1 - 3*b))/(3*(29 + 9*b + 3*Sqrt[3]*Sqrt[31 + 18*b -
b^2 - 4*b^3])^(1/3)) +  (29 + 9*b + 3*Sqrt[3]*Sqrt[31 + 18*b - b^2 -
4*b^3])^(1/3)/(3*2^(1/3))

but I still do not have any idea how to persuade Mathematica to give me
this (or any reasonable) result.

(Regarding Root and ToRadicals, now I understand that Root is numbering
the roots differently for low and high b values:

   Plot[Root[-1 - b*#1 - #1^2 + #1^3 & , 1] , { b, 0, 10} ]

but when I am applying ToRadicals, it freezes the symbolic form of the
root with respect to the low b. Thanks for helping me to understand
that.)

Peter


Andrzej Kozlowski wrote:
> On 13 Sep 2006, at 17:03, p-valko at tamu.edu wrote:
>
> > (Reduce[{z^3-z^2- b z-1==0, b>0, z>0}, z] //ToRadicals)/.b->3.//Chop
> >
> > gives a negativ z:
> >
> > z == - 0.414214
> >
> > What am I doing wrong?
> >
> > Regards
> > Peter
> >
>
> Why do you need to use ToRadicals here? There is no point in it and,
> in fact you, are only asking for trouble. I am not sure why
> Mathematica even tries to convert Root objects with parameters to
> radicals, since it can't be done reliably due to branching problems
> and is of no practical use whatever, but at least the Help tells you
> that:
>
> â?¼ If Root objects in expr contain parameters, ToRadicals[expr] may
> yield a result that is not equal to expr for all values of the
> parameters.
>
> And, of course, if you do not use ToRadicals, there is no problem:
>
> Reduce[{z^3 - z^2 - b*z - 1 == 0, b > 0, z > 0}, z] /. b -> 3
> 
> z == 1 + Sqrt[2]
> 
> 
> Andrzej Kozlowski
> Tokyo, Japan


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