Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54670] Re: [mg54630] finding roots of 1 + 6*x - 8*x^3
  • From: "Kennedy" <kennedy at oldnews.org>
  • Date: Fri, 25 Feb 2005 01:19:37 -0500 (EST)
  • References: <200502240821.DAA13324@smc.vnet.net> <97ce11791d39a4408f41ac3e4e4b8613@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

Thank you for this very informative answer. Obviously, I didn't realize I 
was asking the impossible from Mathematica. The results (in terms of Cos) 
given by ComplexExpand were very helpful.

Also thanks to Mr. Lichtblau et al who also diagnosed my disease, Casus 
Irreducibilis.

Regards,
Kennedy

>>From: "Andrzej Kozlowski" <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net

> On 24 Feb 2005, at 09:21, Kennedy 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?
>>
>> Thanks,
>> Kennedy
>>
>> PS.  Mathematica 4.2
>>
>>
>>
> It is part of basic undergraduate mathematics (or should be!)  that the 
> roots of a polynomial equation of degree 3, even when all three are real, 
> may not be expressible in radicals without using complex numbers. This 
> looks like this kind of case.
>
> Root objects are very useful, in fact much mnore than radicals, since one 
> can perform algebraic operations on them and compute their numerical 
> values to arbitrary precision:
>
>
> Chop[N[Roots[1 + 6*x - 8*x^3 == 0, x], 20]]
>
> x ==
>    0.93969262078590838405410927732473010213`20.1505149978\
> 3199 ||
>   x ==
>    -0.76604444311897803520239265055541055426`20.150514997\
> 831998 ||
>   x ==
>    -0.17364817766693034885171662676932325372`20.150514997\
> 831998
>
> ALternatively, if you prefer an expression that is obviously real but does 
> not use radicals, you can do:
>
>
>
> ComplexExpand /@ (x /. Solve[1 + 6*x - 8*x^3 == 0, x])
>
>
> {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]}
>
>
> Let me restate the point. What you are asking for is, in general, 
> mathematically impossible so it's no use complaining that Mathematica 
> can't do it.
>
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/andrzej/index.html
> http://www.mimuw.edu.pl/~akoz/
> 


  • Prev by Date: Re: Why are permutations duplicated with LexicographicPermutations? How to avoid this?
  • Next by Date: Re: Testing the 'type' of a root returned by Solve
  • Previous by thread: Re: finding roots of 1 + 6*x - 8*x^3
  • Next by thread: Re: finding roots of 1 + 6*x - 8*x^3