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MathGroup Archive 2005

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Re: finding roots of 1 + 6*x - 8*x^3

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54653] Re: finding roots of 1 + 6*x - 8*x^3
  • From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Fri, 25 Feb 2005 01:18:57 -0500 (EST)
  • References: <cvk4nr$dnj$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Kennedy" <kennedy at oldnews.org> wrote:
> Hello All,
>
> I am trying to find the roots of
> 1 + 6*x - 8*x^3.
>
> Roots[1+6*x-8*x^3==0,x] yields this ugly thing:
> (made uglier by my converting to InputForm)
>
> x == ((1 + I*Sqrt[3])/2)^(1/3)/2 +
>    1/(2^(2/3)*(1 + I*Sqrt[3])^(1/3)) ||
>  x == -((1 - I*Sqrt[3])*((1 + I*Sqrt[3])/2)^(1/3))/4 -
>    ((1 + I*Sqrt[3])/2)^(2/3)/2 ||
>  x == -(1 - I*Sqrt[3])/(2*2^(2/3)*(1 + I*Sqrt[3])^
>       (1/3)) - (1 + I*Sqrt[3])^(4/3)/(4*2^(1/3))
>
> This monstrosity is chock full of imaginaries,
> even though I know all three roots are real.
>
> I tried Solve too but got the same thing, except
> in the form of a set of replacements. My guess is
> that Solve just calls Roots when handed a poly-
> nomial.
>
> When I ran the above through FullSimplify, I
> got three "Root" objects, the upshot of which is
> that the roots of the polynomial are indeed the
> Roots of said polynomial. Huh.
>
> What command can I use to get the roots into
> a form that are
> (a)  purely real, and
> (b)  in radical form?

It is impossible to satisfy both (a) and (b). This has been known for
centuries as the _casus irreducibilis_. See
<http://mathworld.wolfram.com/CasusIrreducibilis.html>.

But one may satisfy either (a) or (b).
Mathematica chooses to satisfy (b) but not (a).
Instead, if we wish to satisfy (a) but not (b), then we may write the three
solutions as, say, -Cos[2*Pi/9], -Sin[Pi/18] and Cos[Pi/9].

David Cantrell


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