Talk:Cubic function

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Simpler Formulas for Cubic Roots[edit]

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}


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


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} . — (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. — (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). — (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. :-) — (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 ? — (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... :-) — (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... :-\ — (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). — (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... — (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. — (talk) 08:26, 25 April 2013 (UTC)

Tricky without a proper source. <irony>There must indeed have been a good reason why the wp:RS policy was put in place :-)</irony> - DVdm (talk) 16:17, 24 April 2013 (UTC)
Are you saying neither of the two trustworthy sources I offered are acceptable ? — (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. — (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. — (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... :-( — (talk) 07:39, 25 April 2013 (UTC)
The book may be copyrighted, but the link isn't. We can add it like this[1]
  1. ^ Press, William H.; Vetterling, William T. (1992). Numerical Recipes in Fortran 77: The Art of Scientific Computing. Cambridge University Press. p. 179. ISBN 0-521-43064-X. , Extract of page 179
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. — (talk) 17:01, 24 April 2013 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── I have edited the section for

  • 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.

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)

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). — (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) — (talk) 14:01, 25 April 2013 (UTC)
Yet, I do like it :-) - DVdm (talk) 14:07, 25 April 2013 (UTC)
Quite. :-)— (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... — (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|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)

Comments by an IP user[edit]

Master Lagrange (1736 – 1813) revealed at least two centuries ago primitive third roots of unity in terms of the exponents as the solutions of the equation:

 \zeta ^3-1=0=(\zeta -1)( \zeta ^2+\zeta +1)=( \zeta -1) \left(\zeta  -\frac{i\sqrt3-1}{2}\right) \left(\zeta -\frac{2}{i\sqrt3-1} \right) \text{ yielding}  \sqrt[3]1=\zeta^{0;\pm1}=\left(-\frac{1}{2}+i\frac{\sqrt3}{2}\right)^{0;\pm1}=1;-\frac{1}{2}\pm i\frac{\sqrt3}{2} \text { or in trigonometric notation } \sqrt[3]1=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3}=e^{\frac{2ik\pi}{3}}.

The formula in terms of the coefficients can be obtained by means of the section 3.5 Vieta’s (1540 - 1603) substitution which is properly sourced at the works of Nickalls R. W. D. (see Notes 16.” Viète, Descartes and the cubic equation” and 29. "A new approach to solving the cubic”).

Therefore I see no need essentially the same approach to be repeated hence the sections 2 (along with the image NR. 3), 3.2, 3.3, 3.5, 3.7 and 3.8 can be rearranged and titled at following order:

2 Derivatives, the function flow and reduction to depressed cubic equation

The notation implemented here is geometrically grounded hence the variables are the abscise (xS), ordinate (yS = aq) and slope (mS = ap) at the point of the inflection and Symmetry S(xS,yS) where: x_S= \frac{-b}{3a}, y_S=f(x_S)=a x_S ^3+b x_S ^2+c x_S +d=\frac{2 b^3-9abc+27a^2d}{27a^2}\text{ and } m_S=f'(x_S)=3a x_S ^2+2b x_S +c=\frac{3ac-b^2}{3a}.  \text{Inserting }f^{(3)}( x_S)=6a,f''(x_S)=6ax_S+2b=0,f'(x_S)=m_S\text{ and }f(x_S)=y_S\text{ at Taylor’s series about S we get}  f(x)=\frac{6a}{3!}(x- x_S)^3+ \frac{m_S}{1!}(x-x_S)+ \frac{y_S}{0!}=a(x-x_S)^3+m_S(x-x_S)+y_S\text{ being depressed cubic function in }x-x_S=\frac{3ax+b}{3a}.  \text{ Last two items form }f_t(x) = \tan\sigma(x-x_S)+y_S\text{ which is a tangent of } f(x) \text{ at point S where }\tan\sigma=f'(x_S)=m_S.  \text{See graph of } f(x)=x^3-3x^2-144x+432=(x-1)^3-147(x-1)+286\text{ and } f_t(x)=-147x+433=-147(x-1)+286.

\text{Besides, } f'(x)=3a(x-x_S)^2+m_S=3(x-1)^2-147=0\text{ determines } x_{max;min}=x_S\mp\sqrt{\frac{m_S}{-3a}}=1\mp7\text{ existing if }\frac{m_S}{a}<0.

3 Vietè’s substitution and real solution in terms of the inflection point properties

a(x_0-x_S)^3+m_S(x_0-x_S)+y_S=0\text{ or }(3ax_0+b)^3-3(b^2-3ac)(3ax_0+b)=9abc-2b^3-27a^2d

is depressed cubic equation for which we can find real solution (x0) by means of

 x_0-x_S=z-\frac{m_S}{3az}\text{ being Vieta's substitution for this form of the equation yielding quadratic one in } z^3  (z^3)^2+\frac{y_S}{a}z^3-\left(\frac{m_S}{3a}\right)^3=0\text{ satisfied for }z^3=\frac{-y_S}{2a}+\sqrt{\left(
\frac{-y_S}{2a}\right)^2+\left(\frac{m_S}{3a}\right)^3}  \text{ that after rooting and back-conversion gives}  z= \sqrt[3] {\frac{-y_S}{2a}+\sqrt{\left(
\right)^3}}\text{ and }x_0=x_S+\sqrt[3] {\frac{-y_S}{2a}+\sqrt{\left(
\right)^3}}+\sqrt[3]  {\frac{-y_S}{2a}-\sqrt{\left(
\right)^3}} being real number even if the item under square root is negative as confirmed at section 3.2 below implementing trigonometric notation.

Now is proper moment for introducing any of above quoted expression for primitive third roots of unity where we can choose any set of three consecutive integers –1; 0; +1 or 999; 1000; 1001 but I intercede for 0; 1; 2 in order a coherency with a majority of the sections (3.6, 3.7 and 3.9.1) to be maintained. If so, t1 in section 3.4 Cardano’s method should be replaced by t0 enabling x0 to be real solution within entire article.

3.1 Factorization enabling common formula in terms of the inflection point properties

Annuling the quotient of the differences of ordinates and abscises we get

\frac {a(x-x_S)^3+m_S(x-x_S)+y_S- a(x_0-x_S)^3-m_S(x_0-x_S)-y_S}{ a(x-x_S)- a(x_0-x_S)}=(x-x_S)^2+(x_0-x_S) (x-x_S)+ (x_0-x_S)^2+\frac{m_S}{a}=0

\text{for } x_{1:2}=x_S-\frac {1}{2}(x_0-x_S)\pm i\frac{\sqrt3}{2}\sqrt{(x_0-x_S)^2+\frac{4m_S}{3a}}\text{ that inserting Vieta’s substitution } x_0-x_S=z-\frac{m_S}{3az}\text{ becomes}

x_{1:2}=x_S-\frac{1}{2}\left(z-\frac{m_S}{3az}\right)\pm i\frac{\sqrt3}{2}\left(z+\frac{m_S}{3az}\right)=x_S+\left(-\frac {1}{2}\pm i\frac{\sqrt3}{2}\right)z-\left(-\frac {1}{2}\mp i\frac{\sqrt3}{2}\right)\frac{m_S}{3az}=x_S+e^{\pm\frac{2i\pi}{3}}z- e^{\mp\frac{2i\pi}{3}}\frac{m_S}{3az}.

Hence z is now known all of these three solutions can be merged into an algebraic formula in terms of the inflection point properties

x_k= x_S +e^{\frac{2ik\pi}{3}} \sqrt[3]{\frac{ -y_S }{2a} +\sqrt{\left(\frac{-y_S }{2a}\right)^2+\left(\frac{ m_S }{3a}\right)^3}}+e^{\frac{-2ik\pi}{3}} \sqrt[3]{\frac{ -y_S }{2a} -\sqrt{\left(\frac{- y_S }{2a}\right)^2+\left(\frac{ m_S }{3a}\right)^3}} \text{ for }k=0;1;2.

3.2 Algebraic and trigonometric formula for all of three solutions in terms of the coefficients

In order to shorten the algebraic formula to the width of A4 hard copy after inserting of  x_S= \frac{-b}{3a}, y_S=f\left(\frac{-b}{3a}\right)\text{ and }m_S=f'\left(\frac{-b}{3a}\right)=\frac{3ac-b^2}{3a}\text{ we, taking out }\sqrt[3]{\frac{-y_S}{2a}}=\frac{1}{3a}\sqrt[3]{\frac{9abc-2 b^3-27a^2d}{2}},\text{ get} x_k=\left(\frac{e^{\frac{2ik\pi}{3}}}{3a}\sqrt[3]{1+\sqrt{1-\frac{4(b^2-3ac)^3}{(9abc-2 b^3-27a^2d)^2}}}+\frac{e^{\frac{-2ik\pi}{3}}}{3a}\sqrt[3]{1-\sqrt{1-\frac{4(b^2-3ac)^3}{(9abc-2 b^3-27a^2d)^2}}}\right)\sqrt[3]{\frac{9abc-2 b^3-27a^2d}{2} }-\frac{b}{3a} \text{ which is apparently complex if } 4(b^2-3ac)^3>(9abc-2b^3-27a^2d)^2>0\text{ when the trigonometric notation should be}  \text{  implemented by means of }z=e^{i\theta}\sqrt{b^2-3ac}\text{ and }\frac{9abc-2b^3-27a^2d}{2\sqrt{(b^2-3ac)^3}}=\cos{3\theta}=\frac{e^{3i\theta}+ e^{-3i\theta}}{2}\text{ yielding 3 real solutions:}   x_k =  \frac{\left(e^{i\left(\theta+\frac{ 2k\pi}{ 3}\right)}+e^{-i\left(\theta+\frac{ 2k\pi}{ 3}\right)}\right)\sqrt{b^2-3ac}-b}{3a} = 2\cos\left(\frac{1}{3}\arccos\frac{9abc-2b^3-27a^2d }{2\sqrt{(b^2-3ac)^3}} +\frac{2k\pi}{3}\right)\frac{\sqrt{b^2-3ac}}{3a} -\frac{b}{3a}.

Last five lines are presenting all that the innocent little children should know about the resolving of cubic equation – the memorizing of the items involved is facilitated hence all of three are either equal or proportional to the inflection point properties. See the examples: x_0=\frac{1}{2}\left(\sqrt[3]{7+\sqrt{7^2+1^3}}+\sqrt[3]{7-\sqrt{7^2+1^3}}\right)=\sinh\frac{\operatorname{arsinh}{7}}{3}=1\text{ is real solution of } 4x^3+3x=7\text{ and} x_0=\frac{1}{2}\left(\sqrt[3]{26+\sqrt{26^2-1^3}}+\sqrt[3]{26-\sqrt{26^2-1^3}}\right)=\cosh\frac{\operatorname{arcosh}{26}}{3}=2\text{ is real solution of } 4x^3-3x=26\text{ but} x_k=2*7\cos\left(\frac{1}{3}\arccos{\frac{-286}{2*7^3}+\frac{2k\pi}{3}}\right)+1=12;-12;3\text{ are real solutions of } (x-1)^3-3*7^2(x-1)=-286. Note: Obviously the evaluation of first two examples will be much easier if hyperbolical formulae would be applied.

Comment: It’s incomprehensible that a point of such importance for THE FLOW OF THE FUNCTION isn’t mentioned within entire article although the coordinates and slope of S are determining all remaining characteristic points: the zero(s), critical points, if any, as well. Moreover, the circle at Figure 4 is unnecessary dislocated above S(xS,yS) omitting even a vertical line up to the center of the circle as done at Figure 2 of “New approach …” by Nickalls R. W. D. (see note 29 again). Tschirnhaus transformation isn’t only unneeded but also responsible for such a failure hence t_S = 0 doesn’t mean that an inflection point S(0,q) doesn’t exist. It seems to me reasonable the unknown t along with p and q to be abandoned in favour of z, θ and geometrically grounded variables which are also easier for memorizing.

Note: Letter S (instead N at 29) is chosen not only for Symmetry but also due to its shape reminding to cubic function. (talk) 09:02, 19 July 2013 (UTC)Stap Modified many times by (talk) 23:13, 2 May 2014 (UTC)

Omar Khayyám's solution - image shows special case[edit]

In this image vertical line goes through the center of the circle, but that's not always the case. Less confusing image would have a vertical line not going through the center of the circle. (talk) 09:07, 6 June 2013 (UTC)

It's worse than just confusing, it's misleading. I started to verify it for myself, then thought "Hey! Why bother? He already knows the value, it's just the radius of the circle." I'm going to check it further, and may do something about cleaning things up. Maurice Fox (talk) 23:30, 20 February 2014 (UTC)

Please do! I too was lead astray for a moment, thinking "why not just construct x=y?" All the best: Rich Farmbrough11:01, 27 April 2014 (UTC).

Well, it's worse than misleading, it's just plain wrong. I had forgotten about this comment, but your note inspired me to look it up. In the first place, see the references. One of them speaks about a hyperbola, not a parabola. Anyway, take this counterexample: a = 2, b = 5, which gives x^3 + 4x = 5. Easily, x = 1 is a solution of the equation. The depressed equation is x^2 + x + 5 = 0, which has no real roots. According to the piece, we should get the intersection of the circle (x - 5/4)^2 + y^2 = (5/4)^2 and the parabola y = x^2 / 2. But when x = 1, the parabola y = 1/2 and the circle y = sqrt(3/2). I'm not going to bother with attempting to fix this. Maybe someone should just delete it, or some diligent person should fix it. I guess I'll put a note in the article to save others the frustration. Maurice Fox (talk) 20:48, 27 May 2014 (UTC)

The equation of the circle is (x - 5/8)^2 + y^2 = (5/8)^2 (do not confuse radius and diameter). With the correct equation of the circle, the parabola and the circle intersect at x=1, y=1/2. D.Lazard (talk) 22:33, 27 May 2014 (UTC)
I blush to admit that you are right, I did drop a stitch between the radius and the diameter. I'm editing the article to make it clearer. The accompanying graph still needs to be fixed, though, or note that it shows a special case. Maurice Fox (talk) 14:53, 28 May 2014 (UTC)
IMO your edit is too much related to Cartesian coordinates, which were not invented at Kayyam time. I'll rather define the circle by the end points of a diameter, without referring to its center. I guess that this will also avoid confusion, while remaining, in the modern terminology, as close as possible to the original spirit of the method. I'll also edit the caption to give the values of a, b and of the root. D.Lazard (talk) 15:43, 28 May 2014 (UTC)
Nice fixes! Now I'll let the topic drop. Regards. Maurice Fox (talk) 19:39, 28 May 2014 (UTC)

Using Symmetric Notations for Polynomial Coefficients[edit]

If the polynomial coefficients would have symmetrical notations

a x^3 +\ b x^2 + b' x + a'= 0

then expression of Delta1 would be easier to remember

\Delta_1 = 2b^3 - 9a\left(b b' - 3aa'\right) (talk) 17:45, 14 August 2013 (UTC)

Real roots/real coefficients[edit]

I would like to see a solution for common case where you want to find real roots of a cubic, avoiding the use of complex numbers. -- (talk) 19:56, 9 May 2014 (UTC)

If the cubic equation has only one real root, the given formula gives it without using complex numbers. It there are three roots, it has been proven that every formula in terms of radicals involves non-real radicals for all the roots. D.Lazard (talk) 22:18, 9 May 2014 (UTC)
If you are referring to the "General formula", I don't see how you can easily avoid complex numbers? I want to compute only real roots and ignore the complex roots. In the meantime I worked it out with the help of a textbook. The book gives a simple solution in terms of trigonometric and hyperbolic functions, using several case distinctions. -- (talk) 14:51, 10 May 2014 (UTC)
The trigonometric solution you are referring to is probably the one which is described in section Trigonometric (and hyperbolic) method. D.Lazard (talk) 15:17, 10 May 2014 (UTC)
I think that section is not user-friendly at all. If I see it right, the cases p=0/q=0 are not handled. The cosh/sinh formulas can be used to create the complex solutions too, without complex algebra: Drop factor "-2", then for p<0 cosh(...)±I*sqrt(3)*sinh(...) and for p>0 sinh(...)±I*sqrt(3)*cosh(...). Arguments for sinh/cosh the same as stated for the real solutions in the article.-- (talk) 19:57, 10 May 2014 (UTC)
You say "the cases p=0/q=0 are not handled." The case where p = 0 is trivial, as it reduces to t3 = -q. The case where q = 0 is handled perfectly well by the given formula. (Obviously from a computational point of view, in that case the formula is far from being the simplest method, but that is not the point.) The editor who uses the pseudonym "JamesBWatson" (talk) 10:01, 13 May 2014 (UTC)
Trivial or not (I say not), it is not in there. And I don't want to calculate anything, I want a clear, reliable source. It would be nice to see a clean and short presentation of a solution method at the beginning of the section about roots, avoiding complex arithmetic and historical accounts. I have seen elegant versions of Del Ferros Method and of the Trigonometric Method that would fit. One more thing: The article is called "Cubic function", but it is 90% about cubic roots.-- (talk) 22:56, 15 May 2014 (UTC)
Some more remarks, about the section "Cardanos..." Where it says "The two complex roots are obtained...", the meaning of the sentence is totally unclear. I am not even sure it is correct. The two other solutions t2,t3 should be written out instead of left as an exercise for the reader. It doesn't say what to do with the cube roots. You have to take the real roots, as in the schoolbook definition ((-8)^1/3=-2). Not sure what is meant by "considering the complex cube roots", that sounds wrong. The section goes on to discuss the case discriminant<0, which is confusingly stated as "??? not necessarily positive". The solution is written as "t=...", giving the first impression that there is only one solution. This time you have to consider all three roots of the cube root "u=..." to obtain three real solutions (this information is missing too). While the solution is technically correct, it is useless in practice, since it involves complicated calculations in complex arithmetic. Instead you would use the cos(arccos) method as given in the "Trigonometric" section. -- (talk) 20:16, 19 May 2014 (UTC)
Users 93.220 and 62.220 would like to see simple, single moreover "the users-friendly formula"! Besides, they have a few more or less justified remarks. Perhaps they could find the answers in Section 2 above where all features of cubic function are presented (see Sub-section 2) along with single algebraic formula in terms of coefficients (see Sub-section 3.2) which is, in fact, also converted to trigonometric notation in order "complicated calculations in complex arithmetic" to be avoided . (talk) 14:56, 31 May 2014 (UTC)Stap