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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
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