Finding roots of sextic in radicals



On Friday 25 January 2008, you wrote:
> Have you tried solve? This polynomial is the functional composition of
> lower-degree polys so should easily be handled by solve. No need for
> numerical solutions and identify. Take a look at the doc for
> polydecomp if you're interested.

solve gives long lists of roots involving %i
it's known that the roots of the legendre polynomials are real,
so what i want to see is no %i.

polydecomp is no help either.

Andre

>
>        -s
>
> On 1/25/08, andre maute <andre.maute at gmx.de> wrote:
> > On Thursday 24 January 2008, Andrej Vodopivec wrote:
> > > On Jan 22, 2008 5:04 AM, Jordi Guti?rrez Hermoso <jordigh at gmail.com>
> >
> > wrote:
> > > > If my calculations are correct, the roots of x^6 + 3*x^5 + 6*x^4 +
> > > > 3*x^3 + 9*x +9 should all be expressible by radicals. In fact, they
> > > > are all polynomials in terms of 2^(1/3) and a cube root of unity.
> > > >
> > > > How can I make Maxima tell me what the roots actually are?
> > >
> > > I don't know how to compute the roots in radicals, but I have some
> > > code which can sometimes guess how they look like from their numerical
> > > values:
> > >
> > > (%i1) load(identify)$
> > > (%i2) x^6 + 3*x^5 + 6*x^4 + 3*x^3 + 9*x + 9$
> > > (%i3) allroots(%)$
> > > (%i4) identify(%);
> > > (%o4)
> >
> > [x=(sqrt(3)*%i)/2+(3*2^(1/3)-3/2)/3,x=(3*2^(1/3)-3/2)/3-(sqrt(3)*%i)/2,x=
> >.2
> >
> > >250982321872863*%i+(-3/4^(1/3)-3/2)/3,x=(-3/4^(1/3)-3/2)/3-
> >
> > .2250982321872863*%i,x=1.95714903975616*%i+(-3/4^(1/3)-3/2)/3,x=(-3/4^(1/
> >3)
> >
> > >-3/2)/3-1.95714903975616*%i] (%i5) subst(first(%), %o2), ratsimp; <--
> > > you need to check if we
> > > guessed correctly
> > > (%o5) 0
> > >
> > > After you have two roots written with radicals you should be able to
> > > solve the polynomial of degree 4 to get others.
> > >
> > > identify works mostly like the identify from maple but I think is less
> > > powerfull. If you would find it interesting let me know.
> >
> > Perhaps you could check
> > what your identify function gives for the sixth legendre polynomial.
> >
> > 231*x^6/16-315*x^4/16+105*x^2/16-5/16
> >
> > I have not seen the roots written in radicals, yet.
> >
> > Andre
> >
> > _______________________________________________
> > Maxima mailing list
> > Maxima at math.utexas.edu
> > http://www.math.utexas.edu/mailman/listinfo/maxima