Talk:Cubic function/Archive 5

Error in "Q" in the monic trinomial form
There is an error in the formula for "q" in the monic trinomial form.

The denominator should be 27a^2 instead of 27a^3. please someone correct this. You can see a link here:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Or you can derive your own expression (which I did too) — Preceding unsigned comment added by 140.105.70.136 (talk) 15:08, 25 May 2012 (UTC)
 * The formula is correct. In your link the polynomial in y is not monic, as y3 has a as coefficient. Thus you have to divide by a the equation of the link before comparing. D.Lazard (talk) 17:01, 25 May 2012 (UTC)

Geometric interpretation
You don't need perfect vision to see that the angles &theta; are not the same, or that the three red lines intersection points on the circle do not form an equilateral triangle. Ssscienccce (talk) 10:41, 6 September 2012 (UTC)


 * Good catch. I modified the figure by moving the top red line slightly clockwise to make the angles match, and to make it look more like an equilateral triangle. - DVdm (talk) 11:33, 6 September 2012 (UTC)

In the illustration of Omar Khayyám's geometric solution, the circle is drawn so that the parabola intersects it at the top, putting the vertical line through the center of the circle. This suggests a trivial solution, x=b/(2a^2). It would be better to draw the circle with a different diameter (choose a different value of b) so as not to give this false impression. — Preceding unsigned comment added by 64.95.214.18 (talk) 18:55, 19 December 2012 (UTC)

Alternative trigonometric substitution
Instead of the cosine substitution x = u cos(&theta;), we can also use x = u sin(&theta;). But we must associate the coefficients with the identity
 * $$4\sin^3(\theta)-3\sin(\theta) + \sin(3\theta)=0$$

Bad English and a bigoted insistence to CONTROL a page
To whoever keeps changing my edits back to the original:

My contributions are valid and intelligent. Why do you insist on withholding valid information from the people? You do not own this page, all people are allowed to contribute meaningful information.

And if youre going to change my wording or undo my edits, the least you could do is use proper, intelligible English. — Preceding unsigned comment added by 64.134.140.53 (talk) 23:46, 17 September 2012 (UTC)

Tartaglia's work
Interesting article. Im also interested in the historical component of this article. It speaks of Tartaglia competing with Ferrari and Cardano and Fiore. But Im curious to know what Tartaglias method was, and how was it unique from Ferro's method? — Preceding unsigned comment added by 75.172.58.58 (talk) 09:23, 25 September 2012 (UTC)

The historical component is indeed interesting -- one of my math professors spoke of these "duels" as rowdy public exhibitions with money and prestige on the line. If only the general populace were educated enough to enjoy such intellectual endeavors, we might have decent reality television today. — Preceding unsigned comment added by 71.226.189.240 (talk) 02:16, 29 September 2012 (UTC)

Alternative solution to depressed published in 2010 but with a glitch
The April 2010 issue of Resonance has an alternative solution to a depressed cubic found here http://www.ias.ac.in/resonance/April2010/p347-350.pdf

It involves finding values for u,v,w, so that the cubic is reduced to (x+u)^3 = -v(x+w)^3

Just take the cube root of both sides.

Only glitch is that if in the depressed cubic

x^3 +cx + d= 0 c is -3 times q^2 and d is -2 times q^3 then you get a tautology.

Try it to solve (I have set q = 1) x^3 - 3x - 2 = 0

and you wind up with (x+1)^3 = 1(x+1)^3

which is of no use in finding the solution (x=2 BTW).

Otherwise it works smoothly. (There is also a typo in the artciel on p. 349 where they say v when they mean w in one equation.)--WickerGuy (talk) 20:57, 9 May 2013 (UTC)

Error in the section "Trigonometric (and hyperbolic) method"
After trying to verify the following equation using my calculator and some examples, I find it is not correct.
 * $$t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right) \quad \text{for} \quad k=0,1,2 \,.$$

It should be:
 * $$t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right)-\frac{b}{3a} \quad \text{for} \quad k=0,1,2 \,.$$

So I try to correct it. — Preceding unsigned comment added by Alexvong1995 (talk • contribs) 16:04, 3 April 2013 (UTC)
 * In this section, the cubic is $$x^3+px+q$$. Thus a and b are undefined or, at least, should be taken as 1 and 0 respectively. Thus the previous formula was correct and I have restored it. D.Lazard (talk) 16:50, 3 April 2013 (UTC)

217.197.142.0 (talk) 23:13, 2 May 2014 (UTC)

Simpler Formulas for Cubic Roots
The general cubic equation
 * $$x^3 + a x^2 + b x + c = 0$$

has the roots
 * $$x_i = -{a\over3}+u_i\cdot{t\over3}+\bar u_i\cdot\frac{\Delta_0}{3t}$$

where
 * $$t = \sqrt[3]{\Delta_1 - \sqrt{\Delta_1^2 - 4\Delta_0^3} \over 2}$$ . . . and . . . $$\Delta_1 = 2 a^3 - 9 ab + 27 c$$ . . . and . . . $$\Delta_0 = a^2 - 3  b$$

and
 * $$u_i^3 = -1$$ . are the three complex cube roots of -1, and . $$\bar u_i$$ . are the complex conjugates of . $$u_i$$.
 * $$u_0 = \bar u_0 = -1\ ,\qquad u_1 = \bar u_2 = {1 - i\sqrt{3} \over 2}\ ,\qquad u_2 = \bar u_1 = {1 + i\sqrt{3} \over 2}$$ . — 79.113.230.93 (talk) 04:22, 23 April 2013 (UTC)


 * I have removed some blank lines from your message—hope you don't mind.
 * This will not stand !! :D
 * Very nice. If you have the source (book/journal/article title, author, publisher, date, page...), let's put it in the article. - DVdm (talk) 07:40, 23 April 2013 (UTC)
 * It's the exact same thing as what's already in the article itself, only more "organized" (instead of constantly and tirelessly repeating the same lengthy expressions over and over and over again). It's not a new formula, or anything. I proposed a similar approach for the quartic one as well, also based on what was already there, only "grouped". Maths isn't "chaotic" and "complicated" if you don't want it to be. — 79.113.230.93 (talk) 09:20, 23 April 2013 (UTC)


 * I agree, and it is perfectly OK, but we can't take in the article per wp:NOR and it clearly is not a trivial calculation as in wp:CALC. Bummer. Unless of course we all agree and do the wp:IAR. What say we? - DVdm (talk) 09:36, 23 April 2013 (UTC)
 * It's not a "calculation", let alone "research" (original or otherwise), it's simply a notation. An abbreviation. Nothing more. Like saying "Pi" instead of "3.14...". It doesn't add any "new information", it just simplifies the writing, that's all (and perhaps also facilitates the understanding). — 79.113.230.93 (talk) 09:49, 23 April 2013 (UTC)


 * Yes, I agree. Unless someone objects, feel free to add it to the article. Ditto for quartic. Make sure you mention the talk page in your edit summary. - DVdm (talk) 10:29, 23 April 2013 (UTC)


 * I agree with the IP editor and DVdm. This is nothing but a more concise way of writing the same thing, avoiding, for example, writing the great long square root six times. OK, if there is no reliable source that gives the formula in this precise form then it is possible to argue that it is original research. However, rearranging a formula is really not at a very much higher level of "research" than rephrasing information into one's own words. I see no good reason at all for not including it in the article. I don't regard this as ignoring rules, because I regard it as an unreasonably pedantic interpretation of the "rules" to exclude this as "original research": it is not the sort of thing that the policy on original "research" is intended to stop. Also, as far as I can see, the formulas for the roots already quoted in the article are not referenced to any source, so if we are going to go in for the line that no formula can be quoted in the article without a source then the whole lot will have to be removed. Just stick the clearer and simpler version in, say I. JamesBWatson (talk) 10:37, 23 April 2013 (UTC)


 * I already went ahead at quartic. I'll leave the one here for IP79 or JamesB :-) - DVdm (talk) 11:34, 23 April 2013 (UTC)
 * Please note that a formula which is very similar to the above one appears already at the end of section "General formula for the roots. D.Lazard (talk) 12:42, 23 April 2013 (UTC)
 * LOL! You guys were actually right: "my" formula was indeed a bit different... Apart from having the coeficients normalized, it also uses the fact that $$t \cdot \bar t = \sqrt[3]{2} (a^2 - 3 b)$$, where $$\bar t = \sqrt[3]{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3} \over 2}$$ . So I chose to "modify the modification", and instead just simplify what was already there ! Hope everyone is satisfied now. :-) — 79.113.213.225 (talk) 07:04, 24 April 2013 (UTC)
 * Sorry, but the notation using a double index for two simultaneous definitions and the sophisticated use of \pm and \mp are not convenient for most readers. D.Lazard (talk) 07:33, 24 April 2013 (UTC)
 * Better now ? — 79.113.213.225 (talk) 07:39, 24 April 2013 (UTC)


 * Yes. I can live with it. But I am not sure to prefer this version to the old one with the formula expanded: The new version somewhat masks the true complexity of the formula. For a beginner, and in particular for a kid, it may be interesting to know that there exist big formulas and to understand (without any explicit explanation) why they spent so much time on quadratic equation and not on cubic equation. This is for this kind of reasons that, with computer algebra systems, many beginners ask for the solutions of the general cubic equation. The opinion of other editors would be welcome on this point. D.Lazard (talk) 08:30, 24 April 2013 (UTC)
 * That's sort of like saying that we should intentionally write all numbers on Wikipedia in Roman numerals, to make people better appreciate the usefulness of the Arab ones, which they might otherwise take for granted... :-) — 79.113.213.225 (talk) 09:13, 24 April 2013 (UTC)

Something's wrong. Try x^3 + x = 0 with (x0,x1,x2) = (0,i,-i). The scheme here above with (a,b,c) = (0,1,0) produces the correct result. The current article scheme with (a,b,c,d) = (1,0,1,0) does not. Work ahead. Where is that source? - DVdm (talk) 10:42, 24 April 2013 (UTC)
 * I do not understand your point: the cubic roots are 3 and -3, which produces the correct result. D.Lazard (talk) 12:03, 24 April 2013 (UTC)
 * The old formula was indeed incorrect... Not only that, but so was another one I've found online. WTH ? This is getting really annoying, and proving to be significantly harder than I initially thought... :-\ — 79.113.213.225 (talk) 12:33, 24 April 2013 (UTC)
 * Unless either this or this might count as possibly valid links. (It's correct, alright, but I'm not so sure if it's OK to copy their content elsewhere). — 79.113.213.225 (talk) 13:12, 24 April 2013 (UTC)
 * IMO, all formulas in the article were correct except the present one: Presently, you divide twice by 3a, inside the radicals and at the end. The resulting formula is thus homogeneous of degree -1 in a,b,c,d, when it should be of degree 0. D.Lazard (talk) 13:28, 24 April 2013 (UTC)
 * In Mathematica's opinion, they were wrong. NOW the formula is finally correct, but sourcing it seems to be a problem... Let's just hope that we can use at least one of the two (correct) links provided above... — 79.113.213.225 (talk) 13:36, 24 April 2013 (UTC)


 * Yes, the current version is correct. The previous version was not, as with my example x*(x+i)*(x-i) = x^3+x with a=c=1 and b=d=0 it produced x0=-(√3+i)/2, x1=0, x2=(√3+i)/2. The current version and scheme here above on the talk page produce x0=0, x1=i, x2=-i. Tricky business without sources ;-)
 * By the way, I changed {0,1,2} to {1,2,3} for consistency within the article. I made another little tweak on the quartic article. - DVdm (talk) 15:43, 24 April 2013 (UTC)


 * Sorry, but I disagree with the present version with a single cubic root appearing twice. The formula with two different cubic roots appears in many (maybe most) textbooks and was Cardano's formula. It has thus to appear first. The fact that it is wrong when the square root is not real is an important fact, that is explained in the article, just after this formula. This is the reason why Mathematica (and also Maple) can not simplify to zero when substituting in the equation: their simplifiers make not the distinction between real and complex. I do not know which formula is provided in Mathematica, but until Maple V, the Maple's solve function provided the formula with two cubic roots and could not simplify to zero the substitution of the solution in the equation. This was corrected in version V.5 only. This shows how this error is common, even among experienced mathematicians. In any case, if the version with a single cubic root remains at the beginning, the whole section must to be rewritten, and I do not see how to do this, by keeping this important warning in an encyclopedic style. D.Lazard (talk) 16:10, 24 April 2013 (UTC)

D.Lazard, I think I found a way to do just that: Please take a look again at the article, and tell me if you're satisfied. — 79.113.215.181 (talk) 08:26, 25 April 2013 (UTC)
 * Tricky without a proper source. There must indeed have been a good reason why the wp:RS policy was put in place :-) - DVdm (talk) 16:17, 24 April 2013 (UTC)
 * Are you saying neither of the two trustworthy sources I offered are acceptable ? — 79.113.213.225 (talk) 17:01, 24 April 2013 (UTC)
 * Not to me. Dr. Math is anonymous and i.m.o. certainly does not comply with wp:RS - check that policy and then read their about. The Wolfram site is a commercial vehicle for Mathecatica and is full of errors. As an example, here (and here) is one glaringly obvious error. It has been sitting there since ages. I sent a first email to Dear Dr. Weisstein in October 2005, and four more —all friendly— emails later. Never even got a reply. They got it right here, but the other junk is still sitting there. No, I don't trust anything on these sites, but some mileages seem to vary here ;-)
 * Now, isn't there any good proper old-fashioned book (that thing made of paper) that list these formulae? - DVdm (talk) 17:40, 24 April 2013 (UTC)
 * Billions! :-) But I'm from Romania... perhaps an Anglo-Saxon bibliography would look better. Besides, the things that Weisstein got wrong were written, not calculated, as the link I provided. BTW, there's an apparently famous Anglo-Saxon mathematical work, an anthology, huge, full with almost every formula, but I don't know its name: maybe that might hopefully help. — 79.113.213.225 (talk) 18:15, 24 April 2013 (UTC)
 * Is this better ? It looks like it's from a published book... He mentions or references a work of Viete, De emendatione, but it's hard to locate the text within the book itself. — 79.113.213.225 (talk) 18:59, 24 April 2013 (UTC)
 * From the Max-Planck-Institut. Looks good for me...- DVdm (talk) 21:25, 24 April 2013 (UTC)
 * But it seems copyrighted... :-( — 79.113.215.181 (talk) 07:39, 25 April 2013 (UTC)
 * The book may be copyrighted, but the link isn't. We can add it like this


 * Nice. - DVdm (talk) 07:56, 25 April 2013 (UTC)
 * Nice. - DVdm (talk) 07:56, 25 April 2013 (UTC)


 * 1. If someone wants to mention Cardano and his endeavours, I think it's best to put those in the History section of the article, as opposed to the General Formula section (since obviously his formula is faulty or incomplete). Obviously people going to the latter section (say pupils or students trying to resolve a cubic for homework) are searching for the best and more accurate expression, not for the first historically.
 * 2. Regardless of WHICH formulas you may ultimately choose to use, please try to keep them as brief andf succint as this one.
 * That would be all. — 79.113.213.225 (talk) 17:01, 24 April 2013 (UTC)

I have edited the section for I am not fully satisfied by the following minor fact: The definition of C, which is the most important part of the formula is visually too far from the solution itself. D.Lazard (talk) 11:07, 25 April 2013 (UTC)
 * Giving a more symmetric shape to the formula
 * Naming C the cubic root (initial of "cubic", capitalized to emphasize that it is a complex expression)
 * Removing duplicates
 * Correcting some errors (for example in the reference to the discriminant)
 * Adding some comments on the choice or the square and cubic roots and its relation to the numbering of the solutions.
 * You beat me to it! :-) I've just now opened up my computer because I've realized that both formulas are "faulty", one when t = 0, and the other one when Delta < 0. When t = 0, the first one becomes the second, that of Cardano, but even this one cracks down when Delta < 0. At least now the article is finally complete and well-written (well-thought and well-organized: logically, as well as visually and aesthetically). — 79.113.221.50 (talk) 13:35, 25 April 2013 (UTC)


 * In four words: nice! - DVdm (talk) 13:50, 25 April 2013 (UTC)
 * Uhm... No, DVdm... it kinda isn't! :-| And do you know why ? :-) Because the exact same problems translate themselves over here... :D And, as you can see, those formulas are even more complex, meaning even more special cases... (evil sadistic laughter) — 79.113.221.50 (talk) 14:01, 25 April 2013 (UTC)
 * Yet, I do like it :-) - DVdm (talk) 14:07, 25 April 2013 (UTC)
 * Quite. :-)— 79.113.221.50 (talk) 14:15, 25 April 2013 (UTC)
 * D.Lazard, I also disagree with you on WHY cubics aren't taught to innocent little children: It's NOT because "when we write the roots DIRECTLY in terms of coefficients, those formulas are SOOO looong!"... but rather it's because of the complex cases and sub-cases of all sorts of special values for all discriminants involved... which in the case of the humble quadratics simply do not exist: You extract a complex root of a negative number, but that's it: threre are no real "choices" and "discussions" there... — 79.113.221.50 (talk) 14:01, 25 April 2013 (UTC)
 * For the current general formula, I think we could get rid of $$u_i$$ and simplify a bit by replacing i with n and writing $$(\frac{-1 + i\sqrt{3}}{2})^n$$ and $$(\frac{-1 + i\sqrt{3}}{2})^n$$. The three possible values of n give 1 and the two possible values of ui. Pokajanje &#124; Talk  04:24, 9 May 2013 (UTC)


 * I guess that you mean to replace $$u_i$$ by $$(\frac{-1 + i\sqrt{3}}{2})^i$$. This could be done, but I am not sure it is a good idea: this will make less clear that when $$u_i=1$$, the cubic roots of unit do not appear in the formula. D.Lazard (talk) 09:07, 9 May 2013 (UTC)