Talk:Cubic equation

There are two examples on the spanish version. The method is slightly different. If you copy them, please quote the author.

Accuracy Dispute

This page doesn't seem to be correct. Consider the equation:

<math>x^3 - x = x (x^2 - 1) = x (x - 1) (x + 1)<math>

This has three distinct real roots (0, 1, and -1). However, according to the page, it has <math>a_0 = 1<math>, <math>a_2 = -1<math>, and <math>a_1=a_3 = 0<math>, and thus <math>q=4<math> and <math>r=1<math> and thus <math>s=4<math> and the equation should have "two real roots, one of which is a double root."

It may be that the problem is just in the definition of the cubic — it is usually the convention that <math>a_n<math> is the coefficient of <math>x^n<math> (the article reverses the order). However, since obviously no one has ever checked this article, I can't be confident in any of the other equations. (And I don't have time now to check them myself.)

—Steven G. Johnson

This accuracy dispute has been sitting here for over a month with no activity related to it either here or on the main article. Is anyone paying attention to it? Bryan 08:22, 21 Jan 2005 (UTC)

I happened to stumble on the summary of Bryan's edit. The original poster is right in that there is a mistake in the article. -- Jitse Niesen 19:48, 21 Jan 2005 (UTC)

It turned out that the correct definition for t, mentioned in the previous article, is

<math> t = \frac{(2a_1^3-9a_0a_1a_2+27a_0^2a_3)^2}{(a_1^2-3a_0a_2)^3} <math>

instead of

<math> t = \frac{(2a_2^3 - 9a_1a_2a_3 + 27a_0^2a_3)^2}{(a_2^2 - 3a_1a_3)^3}. <math>

So Steven was right that part of the problem is the definition of the cubic. But even after fixing this there is still the issue of what happens if the denominator vanishes. So I used the discriminant instead.

This would not have been a problem if someone had not changed the cubic polynomial back to <math>a_3x^3 + a_2 x^2 + a_1 x + a_0<math> without changing the definition of t. This is not the standard way to write a cubic polynomial in the theory of equations or algebra in general. This is because if you write it the other way, the coefficents <math>a_n<math> become homogenous polynomials of degree n in the roots. You can see in the above expressions for t that in the first form, the numerator and denominator are both homogenous of degree six, and hence t has weight zero in the roots. This fact is not apparent in the second form.

I'm inclined to change it back to conform to the usage of mathematicians who work in this area. Any comment? Gene Ward Smith 22:04, 23 Jan 2005 (UTC)

I also added a short section on history; more can be written here as it is a nice story. I also removed the paragraph on which branch of the cube root to take, as I couldn't see its relevance. Furthermore, I made some more chances throughout and checked everything, except for (5) and the section on Chebyshev radicals. However, I still have a couple of problems with the page:

• I don't understand the section Factorization. It states that we can find the real root by extracting cube roots only of positive quantities, but how?
• The definition of <math>C_{\frac13}<math> is confusing: we have <math>C_{\frac13}(0) = 2 \operatorname{arccosh} \frac13<math> which is not a real number, but both the power series and the statement that this function solves <math>x^3-3x=t<math> suggest that it should be real. Furthermore, the definition of p in this section differs from the p in the rest. Does somebody have a reference for these Chebyshev radicals?

If we have one real root, we can find it by extracting the cube roots of real numbers. If for some reason we wanted only to take roots of positive numbers, and had negative ones, we could substitute -x for x, find the roots of that, and take the negative of the result. Gene Ward Smith 22:17, 23 Jan 2005 (UTC)

I understand that. The problem is, how do we get the number from which we need to take the cube root? I took the text in the article to mean that there is a way to do this, but I am not sure what formula archieves this. -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

The definition given is in terms of the power series and its analytic continuation, and the power series gives <math>\sqrt{3}<math> for <math>C_{\frac13}(0)<math>. I think I'll change it so that it is clearer that the definition is not in terms of the arccosh function, but it says so as it stands. The equations given for the roots give <math>\sqrt{3}<math>, <math>-\sqrt{3}<math>, and 0 for the three roots, which is correct. Gene Ward Smith 22:04, 23 Jan 2005 (UTC)

I still don't understand what the cosh is doing there. Are you sure it shouldn't be cos? That would make everything real on the interval [-2,2]. And please, do give a reference; it would improve the article and be a big help to us checking it (at least personally, I find I often make mistakes with cubic and quartic equations!). -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

It is possible to use either cosh or cos; it's essentially equivalent. Using cos, we can see this relates to the solution of the cubic using trigonometic functions; however putting it that way disguises the fact that it really comes down to an entirely algebraic solution, in terms of an algebraic function, and hence is exactly analogous to using radicals.

Here's an explanation of the trigonometric method:

http://www.sosmath.com/algebra/factor/fac111/fac111.html

Here's a discussion of Omar Khayyam's geometric method, which is closely related to this business:

http://jwilson.coe.uga.edu/emt669/Student.Folders/Jones.June/omar/omarpaper.html

There's a long and convoluted history involving this approach. So far as the mathematical accuracy of what I wrote, I'm a mathematician who specializes in this area, so I come in that way self-referenced. Gene Ward Smith 21:52, 24 Jan 2005 (UTC)

So I didn't remove the accuracy dispute tag. -- Jitse Niesen 19:55, 23 Jan 2005 (UTC)

Continued accuracy dispute

I notice that the disputed tag was removed. However, the page still isn't correct for the simple example I gave above:

<math>x^3 - x = x (x^2 - 1) = x (x - 1) (x + 1)<math>

which has roots 0, -1, and 1. According to the revised page's notation, this has: <math>\alpha_3 = 1<math>, <math>\alpha_1 = -1<math>, and <math>\alpha_2 = \alpha_0 = 0<math>. Then, according to the page, <math>\Delta = -4 < 0<math>, and the equation should have "one real roots and a pair of complex conjugate roots."

That's my mistake. I was confused by the fact that MathWorld uses different sign conventions for the discriminant on their web site. Sorry. -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

Please don't remove the disputed tag until you at least check all the equations against the roots of a few simple examples. (I don't care what convention you use for the coefficient numbering, but please be consistent!) —Steven G. Johnson 04:31, Jan 24, 2005 (UTC)

What would you say to my putting the coefficients the other way around, to correspond to how algebraists most often write them, putting the sign of the discriminant to be what algebraists usually want, checking everything, and removing the disputed tag? Gene Ward Smith 00:03, 25 Jan 2005 (UTC)

Please go ahead. I'm a bit surprised that you want to put the coefficients the other way around, but I'm sure you know best. If you prefer to use the t from Chebyshev radicals in the first section instead of the determinant, that's fine with me as well. When I asked you to provide some references, I meant the old-fashioned kind: books and papers. I did not want to question your knowledge, it's just that I'm a mathematician myself and I have often quite a bit of trouble getting all the signs right; I know many of my colleagues have similar problems and I presumed the same goes for you. I hope this clarifies the matter. -- Jitse Niesen 13:10, 25 Jan 2005 (UTC)

I've removed the disputed tag since I've checked at least through the Cardano's method section for a few sample polynomials and it seems to work (modulo a couple of special cases which I've now noted in the article). —Steven G. Johnson 19:53, Jun 5, 2005 (UTC)

Complete formula

For the form:

x³ + bx² + cx + d = 0.

<math>x=\frac{-\Bigg(\sqrt[3]{(6\sqrt{3(4b^3d-b^2c^2-18bcd+4c^3+27d^2)}+4b^3-18bc+54d) ^2}+2b\sqrt[3]{3\sqrt{3(4b^3d-b^2c^2-18bcd+4c^3+27d^2)}+2b^3-9bc+27d} +2(b^2-3c)\sqrt[3]{2}\Bigg)}{\sqrt[3]{3\sqrt{3(4b^3d-b^2c^2-18bcd+4c^3+27d^2)}+2b^3-9bc+27d}}<math>

For the form:

ax³ + bx² + cx + d = 0.

<math>x=\frac{-\Bigg(\sqrt[3]{(6\sqrt{3(\frac{4b^3d-b^2c^2}{a^4}-\frac{18bcd+4c^3}{a^3}+\frac{27d^2}{a^2})}+\frac{4b^3}{a^3}-\frac{18bc}{a^2}+\frac{54d}{a}) ^2}+\frac{2b}{a}\sqrt[3]{3\sqrt{3(\frac{4b^3d}{a^4}-\frac{b^2c^2}{a^4}-\frac{18bcd}{a^3}+\frac{4c^3}{a^3}+\frac{27d^2}{a^2})}+2\frac{b^3}{a^3}-\frac{9bc}{a^2}+\frac{27d}{a}} +2(\frac{b^2}{a^2}-\frac{3c}{a})\sqrt[3]{2}\Bigg)} {6\sqrt[3]{3\sqrt{3(\frac{4b^3d}{a^4}-\frac{b^2c^2}{a^4}-\frac{18bcd}{a^3}+\frac{4c^3}{a^3}+\frac{27d^2}{a^2})}+\frac{2b^3}{a^3}-\frac{9bc}{a^2}+\frac{27d}{a}}}<math>

These were based on the formulas given on the page and were tested on a Ti-89... They seemed to work, however there may have been typos during the tex conversion process ... 71.0.202.231 21:38, 4 Apr 2005 (UTC)

Great! I would think these would be too complicated to insert in the article. Wonder what others think. Oleg Alexandrov 23:01, 4 Apr 2005 (UTC)

I don't think it's too useful in practice unless you eliminate the common subexpressions. —Steven G. Johnson 05:51, Apr 5, 2005 (UTC)