Talk:Cubic function/Archive 6

Comments by an IP user

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:

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

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 (x_S), ordinate (y_S = aq) and slope (m_S = ap) at the point of the inflection and Symmetry S(x_S,y_S) where: $x_{S}={\frac {-b}{3a}},y_{S}=f(x_{S})=ax_{S}^{3}+bx_{S}^{2}+cx_{S}+d={\frac {2b^{3}-9abc+27a^{2}d}{27a^{2}}}{\text{ and }}m_{S}=f'(x_{S})=3ax_{S}^{2}+2bx_{S}+c={\frac {3ac-b^{2}}{3a}}.$ ${\text{Inserting }}f^{(3)}(x_{S})=6a,f''(x_{S})=6ax_{S}+2b=0,f'(x_{S})=m_{S}{\text{ and }}f(x_{S})=y_{S}{\text{ at Taylor’s series about S we get}}$ $f(x)={\frac {6a}{3!}}(x-x_{S})^{3}+{\frac {m_{S}}{1!}}(x-x_{S})+{\frac {y_{S}}{0!}}=a(x-x_{S})^{3}+m_{S}(x-x_{S})+y_{S}{\text{ being depressed cubic function in }}x-x_{S}={\frac {3ax+b}{3a}}.$ ${\text{ Last two items form }}f_{t}(x)=\tan \sigma (x-x_{S})+y_{S}{\text{ which is a tangent of }}f(x){\text{ at point S where }}\tan \sigma =f'(x_{S})=m_{S}.$ ${\text{See graph of }}f(x)=x^{3}-3x^{2}-144x+432=(x-1)^{3}-147(x-1)+286{\text{ and }}f_{t}(x)=-147x+433=-147(x-1)+286.$

${\text{Besides, }}f'(x)=3a(x-x_{S})^{2}+m_{S}=3(x-1)^{2}-147=0{\text{ determines }}x_{max;min}=x_{S}\mp {\sqrt {\frac {m_{S}}{-3a}}}=1\mp 7{\text{ existing if }}{\frac {m_{S}}{a}}<0.$

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

$a(x_{0}-x_{S})^{3}+m_{S}(x_{0}-x_{S})+y_{S}=0{\text{ or }}(3ax_{0}+b)^{3}-3(b^{2}-3ac)(3ax_{0}+b)=9abc-2b^{3}-27a^{2}d$

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

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

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

${\frac {a(x-x_{S})^{3}+m_{S}(x-x_{S})+y_{S}-a(x_{0}-x_{S})^{3}-m_{S}(x_{0}-x_{S})-y_{S}}{a(x-x_{S})-a(x_{0}-x_{S})}}=(x-x_{S})^{2}+(x_{0}-x_{S})(x-x_{S})+(x_{0}-x_{S})^{2}+{\frac {m_{S}}{a}}=0$

${\text{for }}x_{1:2}=x_{S}-{\frac {1}{2}}(x_{0}-x_{S})\pm i{\frac {\sqrt {3}}{2}}{\sqrt {(x_{0}-x_{S})^{2}+{\frac {4m_{S}}{3a}}}}{\text{ that inserting Vieta’s substitution }}x_{0}-x_{S}=z-{\frac {m_{S}}{3az}}{\text{ becomes}}$

$x_{1:2}=x_{S}-{\frac {1}{2}}\left(z-{\frac {m_{S}}{3az}}\right)\pm i{\frac {\sqrt {3}}{2}}\left(z+{\frac {m_{S}}{3az}}\right)=x_{S}+\left(-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}\right)z-\left(-{\frac {1}{2}}\mp i{\frac {\sqrt {3}}{2}}\right){\frac {m_{S}}{3az}}=x_{S}+e^{\pm {\frac {2i\pi }{3}}}z-e^{\mp {\frac {2i\pi }{3}}}{\frac {m_{S}}{3az}}.$

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 $x_{S}={\frac {-b}{3a}},y_{S}=f\left({\frac {-b}{3a}}\right){\text{ and }}m_{S}=f'\left({\frac {-b}{3a}}\right)={\frac {3ac-b^{2}}{3a}}{\text{ we, taking out }}{\sqrt[{3}]{\frac {-y_{S}}{2a}}}={\frac {1}{3a}}{\sqrt[{3}]{\frac {9abc-2b^{3}-27a^{2}d}{2}}},{\text{ get}}$ $x_{k}=\left({\frac {e^{\frac {2ik\pi }{3}}}{3a}}{\sqrt[{3}]{1+{\sqrt {1-{\frac {4(b^{2}-3ac)^{3}}{(9abc-2b^{3}-27a^{2}d)^{2}}}}}}}+{\frac {e^{\frac {-2ik\pi }{3}}}{3a}}{\sqrt[{3}]{1-{\sqrt {1-{\frac {4(b^{2}-3ac)^{3}}{(9abc-2b^{3}-27a^{2}d)^{2}}}}}}}\right){\sqrt[{3}]{\frac {9abc-2b^{3}-27a^{2}d}{2}}}-{\frac {b}{3a}}$ ${\text{ which is apparently complex if }}4(b^{2}-3ac)^{3}>(9abc-2b^{3}-27a^{2}d)^{2}>0{\text{ when conversion to polar coordinates should}}$ ${\text{ be applied by means of }}z=e^{i\theta }{\sqrt {b^{2}-3ac}}{\text{ and }}{\frac {9abc-2b^{3}-27a^{2}d}{2{\sqrt {(b^{2}-3ac)^{3}}}}}=\cos {3\theta }={\frac {e^{3i\theta }+e^{-3i\theta }}{2}}{\text{ yielding 3 real solutions:}}$ $x_{k}={\frac {\left(e^{i\left(\theta +{\frac {2k\pi }{3}}\right)}+e^{-i\left(\theta +{\frac {2k\pi }{3}}\right)}\right){\sqrt {b^{2}-3ac}}-b}{3a}}=2\cos \left({\frac {1}{3}}\arccos {\frac {9abc-2b^{3}-27a^{2}d}{2{\sqrt {(b^{2}-3ac)^{3}}}}}+{\frac {2k\pi }{3}}\right){\frac {\sqrt {b^{2}-3ac}}{3a}}-{\frac {b}{3a}}.$

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(x_S,y_S) 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.

217.197.142.0 (talk) 09:02, 19 July 2013 (UTC)Stap Modified many times by 217.197.142.0 (talk) 23:13, 2 May 2014 (UTC)

Omar Khayyám's solution - image shows special case

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. 89.216.19.52 (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 Farmbrough, 11:01, 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

If the polynomial coefficients would have symmetrical notations

$ax^{3}+\ bx^{2}+b'x+a'=0$

then expression of Delta₁ would be easier to remember

$\Delta _{1}=2b^{3}-9a\left(bb'-3aa'\right)$

— 79.113.220.62 (talk) 17:45, 14 August 2013 (UTC)

Real roots/real coefficients

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. --93.220.44.99 (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. --62.226.246.161 (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.--62.226.246.161 (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.--93.220.51.43 (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. --93.220.10.116 (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 . 217.197.142.0 (talk) 14:56, 31 May 2014 (UTC)Stap

"Indeterminate" at Diophantus time

The history section contained "Diophantus found solutions for indeterminate cubic equations in the 3rd century". An editor has recently linked "indeterminate" to Indeterminate (variable). It is clearly wrong, as this article does not contain any definition for "indeterminate equation", and "indeterminate" (as well as variables) were not invented as Diophantus time. This set the question of the meaning of "indeterminate" in this context. One may understand that Diophantus has solved the general cubic equation (the equation with indeterminate coefficients). This is unbelievable several centuries before Cardano. I have therefore replaced "indeterminate" by "various", the most plausible meaning, and tagged it with {{verify source}}. D.Lazard (talk) 20:30, 26 September 2014 (UTC)

Here's a start for a verifiable source:

"Most of the problems solved by Diophantus involve quadratic or cubic equations, usually with one obvious trivial solution." (p. 6)

"Diophantus found rational solutions to cubic equations in what seems to have been essentially this way." (p. 48)

• Mathematics and Its History By John Stillwell

--50.53.45.110 (talk) 01:21, 27 September 2014 (UTC)

Heath discusses "determinate equations" (p. 58ff) and "indeterminate equations" (p. 67ff).

• Diophantus of Alexandria; a study in the history of Greek algebra by Thomas L. Heath

--50.53.45.110 (talk) 02:18, 27 September 2014 (UTC)

Thanks for the reference. Looking on it, it appears that Heath use "indeterminate" in in the sense of undetermined, and that the cubic equations that Diophantus has solved were cubic equations in two variables to be solved in rational numbers (cubic Diophantine equations in two variables). I'll correct the article. However, the article is about "cubic functions", and bivariate equations may hardly be considered as defined by a function. Therefore I am not sure that this reference to Diophantus has its place here. D.Lazard (talk) 09:00, 27 September 2014 (UTC)

Thanks. Cubic equation is a redirect to this article. The cubic equation article was merged with this article. The hatnote and the lead both refer to "cubic equation". --50.53.53.93 (talk) 14:48, 27 September 2014 (UTC)

I know that. But the fact that "cubic function" and "cubic equation" are the same article implies that "multivariate cubic equation" does not belong to the subject of this article. This is what I have tried to explain, and what the hatnote says explicitly. This confirm that the mention to Diophantus in the history section is out of scope. I do not remove it, because I suspect that other sentences of this history section are also out of scope. D.Lazard (talk) 15:42, 27 September 2014 (UTC)

Thanks for your clarification. Heath says: "There is no ground for supposing that Diophantus was acquainted with the algebraical solution of a cubic equation." (p. 66f, bottom) --50.53.53.93 (talk) 16:35, 27 September 2014 (UTC)