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Re: Why is the negative root?

• To: mathgroup at smc.vnet.net
• Subject: [mg69582] Re: Why is the negative root?
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Sat, 16 Sep 2006 03:50:20 -0400 (EDT)
• Organization: The University of Western Australia
• References: <200609130803.EAA18412@smc.vnet.net> <eee0fi$6gl$1@smc.vnet.net>

In article <eee0fi$6gl$1 at smc.vnet.net>, p-valko at tamu.edu wrote:

> "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.)

And indeed, this is what Reduce gives you. The answer given by

  ans = Reduce[{z^3 - z^2 - b z - 1 == 0, b > 0, z > 0}, z]

_is_ the positive root of the equation z^3 - z^2 - b z-1==0 where b>0. 

> 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))

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?

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

Why is the above result, ans, not reasonable or preferable?

In this example,

  Root[#^3 - #^2 - b # - 1 & , 1] // ToRadicals

does give you the radical expression that you are after. However, to 
quote Daniel Lichtblau from TMJ 9(3), here are some reasons to prefer 
the Root form:

[1] It is typically faster to obtain.

[2] It is numerically more stable to evaluate. In general, radical 
formulations are prone to numeric problems. Root objects do not have 
this liability.

[3] When the roots of an irreducible cubic are all real but not 
rational, the so-called "casus irreducibilis" shows that they still must 
be expressed in terms of I. See 

 http://mathworld.wolfram.com/CasusIrreducibilis.html. 

This means that numeric evaluation will give small imaginary parts 
unless, by happenstance, they exactly cancel. Small numeric error from 
round-off makes this unlikely.

[4] For sufficiently complicated algebraics, it is often faster to 
evaluate the Root form numerically, at least at high precision.

[5] Polynomial combinations of Root objects simplify using RootReduce.

[6] Derivatives of Root objects with respect to a parameter are 
expressed in terms of Root objects. This is useful for eigenvalue 
sensitivity analysis.

So, one can avoid the Root form by using ToRadicals -- but for all 
practical computation, you are better off working with Root objects.

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

• References:
  • Why is the negative root?
    • From: p-valko@tamu.edu