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MathGroup Archive 2006

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Re: Why is the negative root?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69642] Re: Why is the negative root?
  • From: p-valko at tamu.edu
  • Date: Tue, 19 Sep 2006 05:44:47 -0400 (EDT)
  • References: <200609130803.EAA18412@smc.vnet.net><eejais$2ta$1@smc.vnet.net> <eel28n$dsf$1@smc.vnet.net>

I am a bit surprised by the "elitism" of the responses. Paul and Andrej
and previously Daniel Lichtbau all defend the Root objects without
telling the whole story. In my opinion those objects are just
pseudo-useful. If you plot
    Plot[Root[-1 + b #1 + #1^3 &, 1], {b,-10,10}]
you will see that they are defending a monster.

But I am not going to start (continue) a debate on faith. Rather I am
trying to formulate my question on a language even hard-core Mathematica 
defenders can accept:

"Assuming that the coefficients are real and I am interested only in
real roots, how do I  persuade Reduce to give the formulas 69 - 72 of

 http://mathworld.wolfram.com/CubicFormula.html ?
(I do not mind if Mathematica gives two different result depending on the sign
of the determinant.)"


Regards
Peter


Paul Abbott wrote:
> > Paul Abbott wrote:
> > > In this case, the single root can be represented by this radical. But
> > > modify your example slightly:
> > >   Reduce[{z^3 - z^2 - b z + 3 == 0, b > 0, z > 0}, z] // FullSimplify
> > > How would you prefer the result to be expressed now?
> >
> > The answer is:
> > b > (-1 - 647/(50867 + 5904*Sqrt[82])^(1/3) + (50867 +
> > 5904*Sqrt[82])^(1/3))/12 &&
> >  ((Sqrt[1 + 3*b] + (2 + 6*b)*Cos[(Pi/2 - ArcTan[(-79 +
> > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/
> >    (3*Sqrt[1 + 3*b]) ||
> >   (Sqrt[1 + 3*b] - (1 + 3*b)*Cos[(Pi/2 - ArcTan[(-79 +
> > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3] +
> >     Sqrt[3]*(1 + 3*b)*Sin[(Pi/2 - ArcTan[(-79 +
> > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/
> >    (3*Sqrt[1 + 3*b]))
> >
> > The answer is in every engineering handbook.
>
> Then every engineering handbook is deficient! Radical formulations are
> prone to numeric problems. Root objects do not have this liability.
>
> Why do you object to Root objects? Is this an "engineering" fetish?
>
> > They call it the "Cardano formula".
>
> I too learnt how to compute the roots of cubics and quartics in high
> school, and I know about the Cardano formula. However, the above
> expression is _not_ the (standard) Cardano formula as it involves trig
> and inverse trig functions. See
>
>   http://mathworld.wolfram.com/CubicFormula.html
>
> Actually, the above expressions are, effectively, Chebyshev radicals:
>
>   http://en.wikipedia.org/wiki/Cubic_equation#Chebyshev_radicals
>
> In general, the Cardono formula is _not_ practically useful. Any
> computation that you need to do involving roots of polynomials is better
> done using Root objects (or using Chebyshev radicals).
>
> Also, consider solving a quintic instead of a quartic ...
>
> Cheers,
> Paul
>
> _______________________________________________________________________
> Paul Abbott                                      Phone:  61 8 6488 2734
> School of Physics, M013                            Fax: +61 8 6488 1014
> The University of Western Australia         (CRICOS Provider No 00126G)
> AUSTRALIA                               http://physics.uwa.edu.au/~paul


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