Re: Re: Why is the negative root?

• To: mathgroup at smc.vnet.net
• Subject: [mg69671] Re: [mg69642] Re: Why is the negative root?
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 20 Sep 2006 02:44:43 -0400 (EDT)

```I also do not wish to enter into useless disputes, however, in my
opinion the essence of this dispute was not about "faith" or even
Mathematica but about mathematics, which by its very nature is an
"elitist" subject. The example you give below illustrates this
perfectly.

Andrzej Kozlowski

On 19 Sep 2006, at 18:44, p-valko at tamu.edu wrote:

> I am a bit surprised by the "elitism" of the responses. Paul and
> Andrej
> and previously Daniel Lichtbau all defend the Root objects without
> telling the whole story. In my opinion those objects are just
> pseudo-useful. If you plot
>     Plot[Root[-1 + b #1 + #1^3 &, 1], {b,-10,10}]
> you will see that they are defending a monster.
>
> But I am not going to start (continue) a debate on faith. Rather I am
> trying to formulate my question on a language even hard-core
> Mathematica
> defenders can accept:
>
> "Assuming that the coefficients are real and I am interested only in
> real roots, how do I  persuade Reduce to give the formulas 69 - 72 of
>
>  http://mathworld.wolfram.com/CubicFormula.html ?
> (I do not mind if Mathematica gives two different result depending
> on the sign
> of the determinant.)"
>
>
> Regards
> Peter
>
>
> Paul Abbott wrote:
>>> Paul Abbott wrote:
>>>> In this case, the single root can be represented by this
>>>>   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?
>>>
>>> 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.
>>
>> Then every engineering handbook is deficient! Radical formulations
>> are
>> prone to numeric problems. Root objects do not have this liability.
>>
>> Why do you object to Root objects? Is this an "engineering" fetish?
>>
>>> They call it the "Cardano formula".
>>
>> I too learnt how to compute the roots of cubics and quartics in high
>> school, and I know about the Cardano formula. However, the above
>> expression is _not_ the (standard) Cardano formula as it involves
>> trig
>> and inverse trig functions. See
>>
>>   http://mathworld.wolfram.com/CubicFormula.html
>>
>> Actually, the above expressions are, effectively, Chebyshev radicals:
>>
>>
>> In general, the Cardono formula is _not_ practically useful. Any
>> computation that you need to do involving roots of polynomials is
>> better
>> done using Root objects (or using Chebyshev radicals).
>>
>> Also, consider solving a quintic instead of a quartic ...
>>
>> Cheers,
>> Paul
>>
>> _____________________________________________________________________
>> __
>> Paul Abbott                                      Phone:  61 8 6488
>> 2734
>> School of Physics, M013                            Fax: +61 8 6488
>> 1014
>> The University of Western Australia         (CRICOS Provider No
>> 00126G)
>> AUSTRALIA                               http://physics.uwa.edu.au/
>> ~paul
>

```

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