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