# [Maxima] Finding roots of sextic in radicals

andre maute andre.maute at gmx.de
Fri Jan 25 08:10:46 CST 2008

```What i did

-------- snip ----------
p : 231*x^6/16-315*x^4/16+105*x^2/16-5/16;

z : solve(p=0,x);

z[1];

z2 : ratsimp(z[1]), algebraic;
z3 : ratsimp(rectform(z[1]));
-------- snip ----------

z1 and z2 contain nevertheless %i.
z3 introduces trigonometric functions
which i dislike as much as the %i

i want an expression only involving radicals,
no %i and no other auxilliary functions

i'm not aware if the trigonometric expressions
have a representation only involving radicals
no %i and no other auxilliary functions

what i have in mind
is something like the following for the fifth legendre polynomial

(%i10) p5:x^5/8-35*x^3/4+15*x/8;
(%o10) 63*x^5/8-35*x^3/4+15*x/8;
(%i11) z5:solve(p5 = 0,x);
(%o11) [x = -sqrt(2*sqrt(70)+35)/(3*sqrt(7)),
x = sqrt(2*sqrt(70)+35)/(3*sqrt(7)),
x = -sqrt(35-2*sqrt(70))/(3*sqrt(7)),
x = sqrt(35-2*sqrt(70))/(3*sqrt(7)),x = 0]

Andre

On Friday 25 January 2008, Stavros Macrakis wrote:
> Did you try radcan and/or ratsimp(zz),algebraic?
>
> On 1/25/08, andre maute <andre.maute at gmx.de> wrote:
> > 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 Gutiérrez 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

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