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Re: Exact real solutions of cubic equations
*To*: mathgroup at smc.vnet.net
*Subject*: [mg46960] Re: Exact real solutions of cubic equations
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Wed, 17 Mar 2004 02:53:40 -0500 (EST)
*Organization*: The University of Western Australia
*References*: <c387p7$bij$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <c387p7$bij$1 at smc.vnet.net>, JonasB at iui.se wrote:
> I would like to find the _exact_ real roots of some cubic polynomials.
> Mathematica seems to have problems determining that a root is real
>
> Solve[1 + a s + b s^2 + s^3 == 0, s]
>
> results in three complex solutions for a = -4 and b = 3. FullSimplify does
> not help, either it does nothing or it gets stuck, depending on the values
> of a and b.
Use ComplexExpand:
Solve[s^3 + 3 s^2 - 4 s + 1 == 0, s] /.
(a_ -> b_) :> a -> ComplexExpand[b] // Simplify
> Does anyone know of a package that can simplify expressions
> with complex numbers?
Physics 52 (3), 269 (1984)), a cubic of the form x^3 - a x + b has
three real roots when a > 0 and 27 b^2 < 4 a^3. In this case, the roots
can be written as
r[n_] = 2 Sqrt[a/3] Cos[Pi (2n - 1)/3 + ArcCos[(b/2)/(a/3)^(3/2)]/3]
where n = 1,2,3. This is easily verified:
Simplify[x^3 - a x + b == 0 /. x -> r[n], Element[n,Integers]]
True
Your cubic reduces to the form x^3 - a x + b by a linear transformation:
Simplify[s^3 + 3 s^2 - 4 s + 1 /. s -> s - 1]
s^3 - 7 s + 7
Clearly a = 7 > 0 and 27 7^2 < 4 7^3 ==> 27 < 28 so the three roots are
2 Sqrt[7/3] Cos[Pi (2n - 1)/3 + ArcCos[(7/2)/(7/3)^(3/2)]/3] - 1
where n = 1,2,3.
However, when all is said and done, I still do not see that these exact
solutions are all that useful. Instead, why not express the solutions
exactly in terms of Root objects
SetOptions[Roots, Cubics -> False, Quartics -> False]
Then,
Solve[1 + a s + b s^2 + s^3 == 0, s]
gives the three roots as exact root objects, as functions of the
parameters a and b. You will see that this representation is useful if
you need to compute a derivate with respect to either or both of these
parameters. For example, computing the partial derivative of the first
root with respect to a,
Simplify[D[Root[#1^3 + b*#1^2 + a*#1 + 1 & , 1], a]]
one obtains a simple function of the root as the result.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 9380 2734
School of Physics, M013 Fax: +61 8 9380 1014
The University of Western Australia (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009 mailto:paul at physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
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