Re: [Maxima-discuss] [Maxima] A few questions about solve and exact polynomial roots

[Stavros M. (Σ. Μ. <mac...@al...>]

It would be really nice to be able to get the trig-form solutions to
polynomials in the main "solve" code, presumably under control of a flag
(solve_trigform?).

           -s

On Wed, Dec 4, 2013 at 11:06 PM, Henry Baker <hb...@pi...> wrote:

> _All_ cubics -- even those with complex coefficients -- can be solved with
> square roots, cube roots and trig functions.  Of course, you don't have to
> use the trig functions, but the solutions are a lot prettier & more
> intuitive if do you use them.
>
> Here's a discussion for the case with 3 real roots.  Almost the same
> derivation works (after shifting & rotating) for the general cubic, so long
> as you realize that the trig functions involved now have general complex
> arguments.  Of course, these general trig functions can be expanded into
> the usual circular & hyperbolic trig functions of real arguments.  I've
> been preparing a similar web page with animations for the general case, but
> it isn't ready yet.
>
> http://home.pipeline.com/~hbaker1/cubic3realroots.htm
>
> At 01:27 PM 12/4/2013, Mike Valenzuela wrote:
> >Wow...
> >
> >Thats so cool. I'm looking forward to the next version (or whichever
> version your function appears)!
> >
> >I'm not sure if there are other causes of casus irreducibilis where trig
> functions work better. I know that *in general* 5th order and higher order
> polynomials do not have closed form solutions in the field of complex
> radicals. I am truly clueless whether there are closed form solutions in
> terms of trigonometric functions. I know for a fact, that many polynomials
> without closed form rational solutions do have trigonometric solutions:
> >x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1, has roots at 2 * cos(2*k*%pi/11),
> for suitable values of k
> >
> >I've also heard of that ultra-radicals one extention to help solve these.
> >
> >I really do thank you for your work as I don't have much time as a
> grad-student. Unless there is a problem so ugly that I cannot do it by hand
> in less time than to code a functional set of rules (and I'm not very
> familiar with the terms I would need to make those rules in the first
> place), then I usually just do the ugly stuff by hand. So I want to thank
> you for your contribution!
> >
> >On Mon, Dec 2, 2013 at 3:03 PM, Aleksas Domarkas <ale...@gm...>
> wrote:
> >Mike Valenzuela on  Dec 2,14:24:50 wrote:
> >
> >>Hello all,
> >>I was messing around with some eigenvectors and values from a simple
> system
> >>and came across the *casus irreduciblis* problem: sometimes complex
> >>radicals are needed to express real solutions. However, I've been trying
> to
> >>reduce the solution to real non-algebraic solutions.
> >>(specified here)
> >>
> http://en.wikipedia.org/wiki/Casus_irreducibilis#Non-algebraic_solution_in_terms_of_real_quantities
> >>(and here)
> >>
> http://en.wikipedia.org/wiki/Cubic_function#Trigonometric_.28and_hyperbolic.29_method
> >
> >>Anyways, I am wondering if maxima natively supports
> >>(1) Returning non-algebraic solutions from its solving system, especially
> >>if they are simplier.
> >>(2) Some combination of trig simplification rules to recover the simple
> >>form given in the above link.
> >
> >>I suppose users could manually use the above identities when applicable,
> >>but they would have to be aware of it.
> >
> > From my "odes" package (to appear in next Maxima relase)
> >function   solvet(eq,x)
> >returns rectform solution of polynomial equation.
> >In "casus irreducibilis" give real solutions expressed
> >in trigonometric functions.
> >
> > Examples:
> >(%i2) solvet(x^3-3*x^2+1,x);
> >(%o2) [x=2*cos(%pi/9)+1,x=2*cos((5*%pi)/9)+1,x=2*cos((7*%pi)/9)+1]
> >
> >(%i3) solvet(x^6-3*x^5-3*x^4+12*x^3-3*x^2-6*x+2=0,x);
> >(%o3)
> [x=1,x=1-sqrt(3),x=sqrt(3)+1,x=2*cos((2*%pi)/9),x=2*cos((4*%pi)/9),x=2*cos((8*%pi)/9)]
> >
> >(%i4) solvet(x^3-15*x-5,x);
> >(%o4)
> [x=2*sqrt(5)*cos(atan(sqrt(19))/3-(2*%pi)/3),x=2*sqrt(5)*cos(atan(sqrt(19))/3+(2*%pi)/3),x=2*sqrt(5)*cos(atan(sqrt(19))/3)]
> >
> >(%i5) solve(x^3-15*x-5,x);
> >(%o5)
> [x=(-(sqrt(3)*%i)/2-1/2)*((5*sqrt(19)*%i)/2+5/2)^(1/3)+(5*((sqrt(3)*%i)/2-1/2))/((5*sqrt(19)*%i)/2+5/2)^(1/3),x=((sqrt(3)*%i)/2-1/2)*((5*sqrt(19)*%i)/2+5/2)^(1/3)+(5*(-(sqrt(3)*%i)/2-1/2))/((5*sqrt(19)*%i)/2+5/2)^(1/3),x=((5*sqrt(19)*%i)/2+5/2)^(1/3)+5/((5*sqrt(19)*%i)/2+5/2)^(1/3)]
> >
> >best
> >
> >Aleksas D
>
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