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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: [mg54655] Re: finding roots of 1 + 6*x - 8*x^3
  • From: Scott Hemphill <hemphill at hemphills.net>
  • Date: Fri, 25 Feb 2005 01:19:01 -0500 (EST)
  • References: <cvk4nr$dnj$1@smc.vnet.net>
  • Reply-to: hemphill at alumni.caltech.edu
  • Sender: owner-wri-mathgroup at wolfram.com

"Kennedy" <kennedy at oldnews.org> writes:

> 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?

What you're asking for isn't possible.  This is a topic for a number theory
course.

  http://mathforum.org/library/drmath/view/53890.html

In[1]:= x /. Solve[1+6*x-8*x^3==0,x]

                            1 + I Sqrt[3] 1/3     1 + I Sqrt[3] 2/3
         -((1 - I Sqrt[3]) (-------------)   )   (-------------)
                                  2                     2
Out[1]= {------------------------------------- - ------------------, 
                           4                             2
 
      1 + I Sqrt[3] 1/3
     (-------------)
            2                        1
>    ------------------ + -----------------------, 
             2             2/3                1/3
                          2    (1 + I Sqrt[3])
 
                                                4/3
         -(1 - I Sqrt[3])        (1 + I Sqrt[3])
>    ------------------------- - ------------------}
        2/3                1/3            1/3
     2 2    (1 + I Sqrt[3])            4 2

In[2]:= ComplexExpand[%]

              Pi                Pi                 Pi                Pi
         -Cos[--]   Sqrt[3] Sin[--]           -Cos[--]   Sqrt[3] Sin[--]
              9                 9        Pi        9                 9
Out[2]= {-------- - ---------------, Cos[--], -------- + ---------------}
            2              2             9       2              2

Is this where you started, trying to find the cosine of Pi/9?

Scott
-- 
Scott Hemphill	hemphill at alumni.caltech.edu
"This isn't flying.  This is falling, with style."  -- Buzz Lightyear


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