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MathGroup Archive 2006

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Re: Why is the negative root?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69627] Re: Why is the negative root?
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Sun, 17 Sep 2006 22:46:11 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <200609130803.EAA18412@smc.vnet.net> <eejais$2ta$1@smc.vnet.net>

In article <eejais$2ta$1 at smc.vnet.net>, p-valko at tamu.edu 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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