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