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Re: Galois resolvent

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96909] Re: [mg96892] Galois resolvent
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Fri, 27 Feb 2009 06:11:16 -0500 (EST)
  • References: <200902261302.IAA26707@smc.vnet.net>
  • Reply-to: drmajorbob at bigfoot.com

So you need these roots?

Solve[t^3 + 2 B t^2 + (B^2 - 4 D) t - C^2 == 0, t]

{{t -> -((2 B)/3) - (2^(1/3) (-B^2 - 12 D))/(
     3 (2 B^3 + 27 C^2 - 72 B D + Sqrt[
        4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
      1/3)) + (2 B^3 + 27 C^2 - 72 B D + Sqrt[
       4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(1/3)/(
     3 2^(1/3))}, {t -> -((2 B)/3) + ((1 + I Sqrt[3]) (-B^2 - 12 D))/(
     3 2^(2/3) (2 B^3 + 27 C^2 - 72 B D + Sqrt[
        4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
      1/3)) - ((1 - I Sqrt[3]) (2 B^3 + 27 C^2 - 72 B D + Sqrt[
        4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(1/3))/(
     6 2^(1/3))}, {t -> -((2 B)/3) + ((1 - I Sqrt[3]) (-B^2 - 12 D))/(
     3 2^(2/3) (2 B^3 + 27 C^2 - 72 B D + Sqrt[
        4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
      1/3)) - ((1 + I Sqrt[3]) (2 B^3 + 27 C^2 - 72 B D + Sqrt[
        4 (-B^2 - 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(1/3))/(
     6 2^(1/3))}}

or these?

Solve[x^4 + B x^2 + C x + D == 0, x] // Simplify

{{x -> (1/(
    2 Sqrt[6]))(\[Sqrt](-4 B + (
         2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
           Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
         1/3) + 2^(
          2/3) (2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3)) - \[Sqrt](-8 B - (
         2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
           Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
         1/3) - 2^(
          2/3) (2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) - (12 Sqrt[6]
             C)/(\[Sqrt](-4 B + (
              2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
                Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                   72 B D)^2])^(1/3) +
              2^(2/3) (2 B^3 + 27 C^2 - 72 B D +
                 Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                    72 B D)^2])^(1/3)))))}, {x -> (1/(
    2 Sqrt[6]))(\[Sqrt](-4 B + (
         2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
           Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
         1/3) +
         2^(2/3) (2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3)) + \[Sqrt](-8 B - (
         2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
           Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
         1/3) - 2^(
          2/3) (2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) - (12 Sqrt[6]
             C)/(\[Sqrt](-4 B + (
              2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
                Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                   72 B D)^2])^(1/3) +
              2^(2/3) (2 B^3 + 27 C^2 - 72 B D +
                 Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                    72 B D)^2])^(1/3)))))}, {x -> -(1/(
     2 Sqrt[6]))(\[Sqrt](-4 B + (
          2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) + 2^(
           2/3) (2 B^3 + 27 C^2 - 72 B D +
             Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
           1/3)) + \[Sqrt](-8 B - (
          2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) - 2^(
           2/3) (2 B^3 + 27 C^2 - 72 B D +
             Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
           1/3) + (12 Sqrt[6]
              C)/(\[Sqrt](-4 B + (
               2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
                 Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                    72 B D)^2])^(1/3) +
               2^(2/3) (2 B^3 + 27 C^2 - 72 B D +
                  Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                     72 B D)^2])^(1/3)))))}, {x -> (1/(
    2 Sqrt[6]))(-\[Sqrt](-4 B + (
          2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) + 2^(
           2/3) (2 B^3 + 27 C^2 - 72 B D +
             Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
           1/3)) + \[Sqrt](-8 B - (
         2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
           Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
         1/3) - 2^(
          2/3) (2 B^3 + 27 C^2 - 72 B D +
            Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 - 72 B D)^2])^(
          1/3) + (12 Sqrt[6]

             C)/(\[Sqrt](-4 B + (
              2 2^(1/3) (B^2 + 12 D))/(2 B^3 + 27 C^2 - 72 B D +
                Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                   72 B D)^2])^(1/3) +
              2^(2/3) (2 B^3 + 27 C^2 - 72 B D +
                 Sqrt[-4 (B^2 + 12 D)^3 + (2 B^3 + 27 C^2 -
                    72 B D)^2])^(1/3)))))}}

Bobby

On Fri, 27 Feb 2009 01:40:41 -0600, Kent Holing <KHO at statoilhydro.com>  
wrote:

> R(t) = t^3 + 2B t^2 + (B^2 - 4D)t - C^2 = 0.
> -----Original Message-----
> From: DrMajorBob [mailto:btreat1 at austin.rr.com]
> Sent: 26. februar 2009 20:21
> To: Kent Holing; mathgroup at smc.vnet.net
> Subject: Re: [mg96892] Galois resolvent
>
> The link includes illegible items such as the question marks in
>
> R(t) = t^3 + 2B t^2 + (B^2 ? 4D)t ? C^2 = 0.
>
> Hence, I have no idea what you're asking.
>
> Bobby
>
> On Thu, 26 Feb 2009 07:02:29 -0600, Kent Holing <KHO at statoil.com> wrote:
>
>> Forthe quartic (*) x^4 + Bx^2 + Cx + D = 0 for integers B, C and D,
>> assume that as for the case C = 0 that all its roots are classically
>> contructible also for the case C /= 0.
>>
>> We can then show that the equation (*) is cyclic (i.e. its Galois
>> group = Z4) iff the splitting field of its Descartes resolvent is E =
>> Q[Sqrt[t0] /= Q for t0 the one and only integer roots t0 of the
>> resolvent. For details, see
>> http://mathforum.org/kb/thread.jspa?threadID=1903146.
>>
>> If the quartic (*) is cyclic, it should be possible using the above to
>
>> explicitly construct the so-called Galois resolvents of (*): The roots
>
>> x1, x2, x3 and x4 of the quartic (*) can be given by polynomials of r
>> with degree less or equal to 3 with rational coefficients for r an
>> arbitrarily root of the quartic. (I.e. the splitting field of the
>> quartic (*) when cyclic is Q[r] for r a root.)
>>
>> Can somebody, using Mathematica, explicitly determine these polynomial
>
>> representations of the roots of the quartic (*). The case C = 0 is
> easy.
>> But the case C /= 0 is indeed messy.
>>
>> Kent Holing,
>> NORWAY
>>
>
>
>
> --
> DrMajorBob at bigfoot.com
>
>
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-- 
DrMajorBob at bigfoot.com


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