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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
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