Re: Re: Why is the negative root?

*To*: mathgroup at smc.vnet.net*Subject*: [mg69615] Re: [mg69608] Re: Why is the negative root?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 17 Sep 2006 22:45:46 -0400 (EDT)

On 17 Sep 2006, at 19:57, p-valko at tamu.edu wrote: > Paul Abbott wrote: >> In this case, the single root can be represented by this radical. But >> modify your example slightly: >> Reduce[{z^3 - z^2 - b z + 3 == 0, b > 0, z > 0}, z] // FullSimplify >> How would you prefer the result to be expressed now? > > The answer is: > b > (-1 - 647/(50867 + 5904*Sqrt[82])^(1/3) + (50867 + > 5904*Sqrt[82])^(1/3))/12 && > ((Sqrt[1 + 3*b] + (2 + 6*b)*Cos[(Pi/2 - ArcTan[(-79 + > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/ > (3*Sqrt[1 + 3*b]) || > (Sqrt[1 + 3*b] - (1 + 3*b)*Cos[(Pi/2 - ArcTan[(-79 + > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3] + > Sqrt[3]*(1 + 3*b)*Sin[(Pi/2 - ArcTan[(-79 + > 9*b)/(3*Sqrt[3]*Sqrt[-231 + 54*b + b^2 + 4*b^3])])/3])/ > (3*Sqrt[1 + 3*b])) > > The answer is in every engineering handbook. They call it the "Cardano > formula". > > Regards, > Peter > I suppose you consider this a "usable formula"? I like your sense of humour. First note that it has LeafCount of 243 vs. 91 for the formula given by Reduce. Moreover, substituting 3 for b into your "engineering Cardano formula" (which is of course not the historical Cardano formula, as that one used only radicals), and applying FullSimplify we obtain the pleasant answer: (1/3)*(1 + 2*Sqrt[10]* Cos[(1/6)*(Pi + 2*ArcTan[13/9])]) || (1/3)*(1 + 2*Sqrt[10]*Sin[(1/3)*ArcTan[13/9]]) while, by contrast, substituting b->3 into the Reduce formula we obtain z == 1 || z == Sqrt[3] If you now apply N to both you will see that they are actually equal. Of course there is no accounting for tastes, but if you really "prefer the first result" I strongly suspect you are in a small minority. Andrzej Kozlowski