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- Thread starter edgo
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CRGreathouse

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If you look at Cardan's formula there are two cube roots taken. Those only drop out if

* the two cube roots are inverses (not sure if this ever happens), or

* both radicands are rational cubes

So it's easy to see that in most cases the solution cannot be written without taking a cube root at some point.

Galois theory is very advanced, you're going to have to take its results on faith for now.

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I am not exactly sure where this all is at, but take the equation X^3=2, of course this can be solved formally in terms of 2^(1/3) + two solutions where we attach the cube root of 1, and then its square, where [tex] \omega = -1/2 +\sqrt-3/2 [/tex]. So clearly, there was never a problem with a formal solution involving a cube root. In fact Galois theory allows the formal taking of nth roots, and I don't think there are any restrictions on that, nor was Galois really concerned with the actual computation.

Of course, in Cardano day imaginary numbers were treated with suspicion and Cardano seems to have equivocated over whether they were of any use or not. They do play an important part in those cubics where there is one real root and two complex, which must be conjugates. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic.

The hard problem was the case of the irreducible cubic, where the solution gives imaginary numbers under the square root, but, in actual fact, has three real solutions, such as X^3-3/4X+1/8. Using [tex] \omega [/tex]one can get a solution for the previous equation, involving the cube root of the cube root of 1. And this can be brought down to earth by using DeMovier’s theorem, arriving at things like the cos(40).

I am not sure how Galois theory impinges on this, since the theory uses only simply arithmetic operations and nth roots. So that if the field was restricted to rational roots, plus adjoining the cube roots of unity, an answer in terms of cosigns would not be forthcoming. But the equation may be considered "solvable "anyway.

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Thanks for your reactions. One might say that I am a believer in Galois; why not, as it is a proved theory. But my issue is a pure theoretical one and is concerning the method of solving a cubic equation. If one wants to solve a quadratic equation, he can’t avoid the presentation of quadratic roots in the solutions. If one wants to solve a cubic equation, Cardano does the job in such a way that it gives cubic roots in the solutions. My question is: is it necessary on theoretical grounds that the solution of the cubic equation comes in the form of cubic roots. That this can result in, let us say

x= ∛8 +∛64, what happens to be 6, is not the problem.

I know one other method, the trigonometric solution of cubic equations but I don’t trust those solutions to be basically free of cubic roots, as they are estimated values that have been given a name.

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CRGreathouse

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My question is: is it necessary on theoretical grounds that the solution of the cubic equation comes in the form of cubic roots.

Cardan's formula, which works for all cubic equations, is in the form

cuberoot(stuff) + cuberoot(other stuff).

So I think the answer to your question is yes. Of course sometimes they can be simplified out (if stuff is 64, say).

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Of course, in Cardano day imaginary numbers were treated with suspicion and Cardano seems to have equivocated over whether they were of any use or not. They do play an important part in those cubics where there is one real root and two complex, which must be conjugates. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic.

This is a really interesting point. In modern education, we introduce the imaginary number [tex]i[/tex] as soon as we start solving quadratic equations. Historically, its use and acceptance came about from its utility when solving

Some days I wonder whether we are cheating our students by teaching mathematics in a different order than it was discovered.

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Look at the equation x^3-3x^2+4X-2. If there is an integral solution is has to divide the constant term, that is, it must be: 1,-1, 2,-2. All we have to do is check for 1, which is a solution. Thus the equation is of the form (X-1)(X^2-2X+2). So all that is now necessary is to solve the quadratic.

However the equation is not irreducible. If it was irreducible over the rationals, then if one root can be found and adjoined to the field, all the roots may be in the extended field. If so, it is called a "Normal Extension"; that's Galois theory.

In the case of a cubic, finding one root and adjoining it to the field reduces the remaining equation to a quadratic. So, assumedly, it would have to be the splitting field. Thus it seems an extension of order 3.

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alxm

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The method of

Poor guy. 500 years later and he's still not getting his dues.. :)

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CRGreathouse

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The method ofTartaglia!

Poor guy. 500 years later and he's still not getting his dues.. :)

Giving Cardan priority follows the standard rule: though del Ferro and Tartaglia developed the method first, they didn't publish.

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It does seem in most cases that the first to publish deserves the credit, since, of course, if someone does not publish at all--why should he later get credit? However there seem to be exceptions to this, like Newton and the Calculus.

I read Cardan's autobiography and I recognize that he fully admitted to the sin of plagiarism after giving his word he would not publish. Yet, mathematicians have always been very kind to this individual, who published much on math and pioneered on probability. Cardan once stated that the only thing he really liked to do was gamble. (This hardly endeared him to respectable society.) In fact, there's a book cheerfully called, "Cardano the Gambling Scholar," by Oystein Ore.

I must add, there is the question of millions of students who take up math, generally as a requirement, and discover that Cardan cheated. This makes it a little tough on the teachers who attempt to prevent cheating on tests.

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Thanks for the answers. Tartaglia: yes, but life isn’t fair. Let us say that Cardano is the product name.

I am trying to understand the Galois related answers, simply because I am not familiar with it. I should be, but it is a long way to go just for being interested in a few answers. What I have been doing is trying to find a basically different way to “treat” cubic equations by embedding them in a structure and trying to kind of solve their structure by means of its variables. A result of that approach is that solving a cubic now can be deducted to solve a pair of cubic equations instead, where the solution is given as a function of two parameters, one for each cubic. This means that solving one cubic in this way can’t be done without solving the other one. Those cubics are linked together by an order-mechanism. I am quite willing to give details but for the moment I want to use the following functions for a last question concerning Galois.

The solution of a cubic with roots x can be given as

x(i) = F * [{4 q(i) - 4 p + 7} /{2 * (4 p -7)^0.5}]

and the other cubic of the pair has roots given by

y(i) = G * [ {4 p(i) - 4 q + 7} /{2 * (4 q -7)^0.5}]

F and G are known functions of the coefficients of the equation to be solved; p and q are variables with the following properties:

If the equation with roots x has coefficients without cubic roots, then p is without cubic roots. If the equation with roots x has roots that can be written without cubic roots, than q is without cubic roots. The equation with roots y is an analogon. This means simply that the solution of an equation that has roots that can be written without cubic roots, comes without cubic roots.

It is proven that the functions that are given above do exist and that the pair of equations that are involved, does exist. Furthermore one can rather easy produce simulated solutions conform the above mentioned functions. The parameter p comes “for free” with the selection of the equation that one wants to solve. The parameter q is easy to locate but I haven’t been able to calculate it in another way than with Cardano, which is the last thing I want to do. My motivation to go on is that q doesn’t fit in cubic roots at all because q is an parameter as p but than for the second equation, being a parameter that is a function of the coefficients of that equation. This function doesn’t create or delete cubic roots, so the cubic roots have to be in the coefficients of the equation. And that equation is, the structure of the system being a mirror, something like a duplicate of the equation that has to be solved and almost sure without cubic roots in the coefficients just like his original (and almost means here: for the time that it is not proved).

If it is dead sure that you can’t make use of the functions given above, because you can’t give the solutions without cubic roots, than I can stop looking for the value of q since it has to be one with cubic roots, though being cubic roots that can be rewritten as functions of quadratic roots. But then I have gone quite a long way to end where I began: an equation with cubic roots that CAN be rewritten as functions with quadratic roots.

That is the reason that I am interested in what can be said about the necessity of cubic roots in the solution of a cubic equation. So my last question regarding this subject is: can it be that the solving of a pair of cubics instead of a single cubic equation interferes in such a way with the Galois Theory that the Theory is not applicable any more for that case?

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So it stops... so a pity.

The other day there was a LaTeX editor available for this forum. I have a new laptop and lost all info about it. Please somebody can give me the necxessary information?

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