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Re: Why is the negative root?

  • To: mathgroup at
  • Subject: [mg69582] Re: Why is the negative root?
  • From: Paul Abbott <paul at>
  • Date: Sat, 16 Sep 2006 03:50:20 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <> <eee0fi$6gl$>

In article <eee0fi$6gl$1 at>, p-valko at 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 

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.


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)    

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