Talk:Cubic function/Archive 6

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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)