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Re: finding roots of 1 + 6*x - 8*x^3
*To*: mathgroup at smc.vnet.net
*Subject*: [mg54681] Re: finding roots of 1 + 6*x - 8*x^3
*From*: Curt Fischer <tentrillion at gmail.NOSPAM.com>
*Date*: Fri, 25 Feb 2005 01:20:35 -0500 (EST)
*References*: <cvk4nr$dnj$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Kennedy wrote:
ComplexExpand works somewhat in your case.
> 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))
In[86]:=
solHmmm = ComplexExpand[x /. Solve[1 + 6*x - 8*x^3 == 0, x]]
Out[86]=
{Cos[Pi/9],
(-(1/2))*Cos[Pi/9] - (1/2)*Sqrt[3]*Sin[Pi/9],
(-(1/2))*Cos[Pi/9] + (1/2)*Sqrt[3]*Sin[Pi/9]}
> What command can I use to get the roots into
> a form that are
> (a) purely real, and
The above expression is purely real, but
> (b) in radical form?
It's not in radical form. How can you represent Sines and Cosines of
Pi/9 in radical form? I'd be interested in a method that does so.
--
Curt Fischer
>
> Thanks,
> Kennedy
>
> PS. Mathematica 4.2
>
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