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
- References:
- Simplify question
- From: "dimitris" <dimmechan@yahoo.com>
- Simplify question