Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*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 2006

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

Search the Archive

Re: Simplify question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71750] Re: [mg71655] Simplify question
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 28 Nov 2006 06:04:04 -0500 (EST)
  • References: <200611260848.DAA14451@smc.vnet.net>

dimitris wrote:
> The following list of expressions was obtained by following the steps
> of Tartaglia's solution of the cubic equation with Mathematica.
> 
> lstcub = {{((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R -
> Sqrt[Q^3 + R^2])^(1/3),
>      -(Q/(R - Sqrt[Q^3 + R^2])^(1/3)) + (R - Sqrt[Q^3 + R^2])^(1/3),
> -(((-1)^(2/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3)) -
>       (-1)^(1/3)*(R - Sqrt[Q^3 + R^2])^(1/3)}, {-(Q/(R + Sqrt[Q^3 +
> R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3),
>      ((-1)^(1/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R +
> Sqrt[Q^3 + R^2])^(1/3),
>      -(((-1)^(2/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3)) - (-1)^(1/3)*(R +
> Sqrt[Q^3 + R^2])^(1/3)}};

This is really unclear. Those are not three cubic roots but rather three 
pairs, presumably taken from two variations of the formulas.


> TableForm[%]//TraditionalForm
> 
> Although the solution of the reduced cubic equation can be obtained
> making a tricky observation
> 
> (see e.g. Leonard E. (Leonard Eugene) Dickson b. 1874. (page 32) ,
> Elementary theory of equations.  1914.
> available online at
> http://mathbooks.library.cornell.edu:8085/Dienst?verb=Display&protocol=CGM&ver=1.0&identifier=cul.math/01460001";
> 
> target=_blank
> 
>>http://mathbooks.library.cornell.edu:8085/Dienst?verb=Display&protocol=CGM&ver=1.0&identifier=cul.math/01460001)
> 
> 
> so what I ask don't play any important role in the solution procedure,
> I wonder if Mathematica can verify the equalities
> 
> MapThread[Equal, lstcub]
> {((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) + (-1)^(2/3)*(R - Sqrt[Q^3
> + R^2])^(1/3) ==
>    -(Q/(R + Sqrt[Q^3 + R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3),
>   -(Q/(R - Sqrt[Q^3 + R^2])^(1/3)) + (R - Sqrt[Q^3 + R^2])^(1/3) ==
> ((-1)^(1/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3) +
>     (-1)^(2/3)*(R + Sqrt[Q^3 + R^2])^(1/3), -(((-1)^(2/3)*Q)/(R -
> Sqrt[Q^3 + R^2])^(1/3)) -
>     (-1)^(1/3)*(R - Sqrt[Q^3 + R^2])^(1/3) == -(((-1)^(2/3)*Q)/(R +
> Sqrt[Q^3 + R^2])^(1/3)) -
>     (-1)^(1/3)*(R + Sqrt[Q^3 + R^2])^(1/3)}
> 
> which can be justified in view of the results
> 
> Table[lstcub /. {R -> Random[], Q -> Random[]}, {5}] // Chop
> 
> Thanks a lot for any response.
> 
> Dimitris

They are not equalities.

{{((-1)^(1/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3) +
    (-1)^(2/3)*(R - Sqrt[Q^3 + R^2])^(1/3),
   -(Q/(R - Sqrt[Q^3 + R^2])^(1/3)) +
    (R - Sqrt[Q^3 + R^2])^(1/3),
   -(((-1)^(2/3)*Q)/(R - Sqrt[Q^3 + R^2])^(1/3)) -
    (-1)^(1/3)*(R - Sqrt[Q^3 + R^2])^(1/3)},
  {-(Q/(R + Sqrt[Q^3 + R^2])^(1/3)) + (R + Sqrt[Q^3 + R^2])^(1/3),
   ((-1)^(1/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3) +
    (-1)^(2/3)*(R + Sqrt[Q^3 + R^2])^(1/3),
   -(((-1)^(2/3)*Q)/(R + Sqrt[Q^3 + R^2])^(1/3)) -
    (-1)^(1/3)*(R + Sqrt[Q^3 + R^2])^(1/3)}}

diffs = Apply[Subtract,lstcub];

First we'll find a set of values for the parameters where the first pair 
of elements clearly differ from one another.

In[6]:= InputForm[val = FindInstance[Abs[First[diffs]]>1, {Q,R}]]
Out[6]//InputForm= {{Q -> -22 + (8*I)/5, R -> 44/5 - (139*I)/10}}

Now we'll look at all pairs evaluated at this set of parameter values. 
We'll see the sets of roots are the same, but the individual pairs are not.

In[8]:= InputForm[N[lstcub/.val,20]]
Out[8]//InputForm=
{{{-0.29403092279589418439472775242462305821`19.922737697936032 +
     0.40042881450871831183411721282038045705`20.05687001240007*I,
    8.27943323194703713516509859094`20.14977275360101 -
     0.48447134348660452129768328543`18.917040238996496*I,
    -7.98540230915114295077037083852`20.15049094673513 +
     0.08404252897788620946356607262`18.172693258269817*I},
   {8.27943323194703713516509859094`20.14977275360101 -
     0.48447134348660452129768328543`18.917040238996496*I,
    -7.98540230915114295077037083852`20.15049094673513 +
     0.08404252897788620946356607262`18.172693258269817*I,
    -0.29403092279589418439472775242464515418`19.922737697936032 +
     0.40042881450871831183411721282041241182`20.05687001240007*I}}}


Daniel Lichtblau
Wolfram Research


  • Prev by Date: Re: sum of integrals over patial intervals != integral over whole interval
  • Next by Date: Re: Re: a technique for options
  • Previous by thread: Re: Simplify question
  • Next by thread: Word permutations - frustrated by lists.