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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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