Talk:Cubic function/Archive 1

Root-finding Formula
When I merged from [ [ Cubic equation]|talk], I noticed each had a different set of formulae for finding the roots. I chose the one that looked simpler, but thought it best to preserve this one as well (just in case the simpler one is wrong or something!) See below: —Celtic Minstrel (talk &#x2022; contribs) 20:08, 6 December 2007 (UTC)

Inconsistency in Cardano's method
It is said that
 * $$ u^{3}={q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}$$
 * $$ u=\sqrt[3]{{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}. \quad (4) $$

But according to the equation some lines lower
 * $$ u=\sqrt[3]{-{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$$

so
 * $$ u^{3}=-{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}$$

I think the latter is consistent with the earlier stated $$y^6+q y^3-{p^3\over 27}=0$$ when applying the quadratic ABC-formula $$y^3={-B\over 2A} \pm \sqrt{\left({B \over 2A}\right)^2-{C \over A}}$$

another inconsistency here
A significant problem it seems to me is that at the introduction (top of page) we have the usual a,b,c,d form of the function. However, at the begining of the Cardano part, we start with a monic form (thats OK) BUT with coefficients 1,a,b,c !!! this is a really bad idea - can we please keep the coeffs a,b,c,d, to `always' represent the original coeffs at the top of the page. To do otherwise is really confusing. Halothane (talk) 18:36, 30 July 2008 (UTC)

another inconsistency here
When discriminant Δ <= 0, ρ = sqrt(-q^3). This does not check the condition that q might be positive resulting in complex ρ. ρ is later used as magnitude in polar form to compute s & t. Complex magnitude is nonsensical. Jhshukla (talk) 18:51, 6 December 2009 (UTC)
 * If the determinant is negative then q is negative. The determinant is q cubed plus a square which must be positive. Dmcq (talk) 19:30, 6 December 2009 (UTC)

The formula
If we have
 * $$ax^3 + bx^2 + cx + d = a(x - x_1)(x - x_2)(x - x_3) = 0,\,$$

let


 * $$q = \frac{9abc - 27a^2d - 2b^3}{54a^3}$$

and


 * $$r = \sqrt{\left (\frac{3ac-b^2}{9a^2}\right )^3 + q^2} = \sqrt{\frac{\Delta}{108a^4}}$$

where Δ is the discriminant defined above. Now, let


 * $$s = \sqrt[3]{q + r}$$

and


 * $$t = \sqrt[3]{q - r}.$$

The solutions are


 * $$x_1 = s+t-\frac{b}{3a},$$


 * $$x_2=-\frac{1}{2}(s+t)-\frac{b}{3a}+\frac{\sqrt{3}}{2}(s-t)i,$$


 * $$x_3=-\frac{1}{2}(s+t)-\frac{b}{3a}-\frac{\sqrt{3}}{2}(s-t)i.$$

Correctness of the formulas
Some preceding comments have noticed inconsistencies in the formulas. It seems that this comes from the imprecision of the definition of cubic roots. I have corrected this in the general formula but not in the remaining of the article. This has yet to be done. This needs some work, and several decisions: one may either change many formulas or precise, when needed, that the formula is correct only in the case of a single real root. This is meaningful as Cardano itself was unable to solve the totally real case which implies to work with non real numbers. Note that a similar correction is probably needed in the page of quartic function D.Lazard (talk) 15:52, 19 May 2010 (UTC)

Merge
I haven't seen any discussion of the merge and so this may be a moot point that has already been demonstrated, but I would just like to point out the following inconsistency:

Linear equation

Quadratic equation

Cubic function

Quartic equation

Qunitic equation

Is there any reason why cubics should be different? asyndeton  talk  20:19, 6 December 2007 (UTC)


 * Yes, I agree that is a little odd. This was rather a hard decision. Switch it around if you like, but it would require a little fiddling with the intro section (ie before the table of contents). The reason I chose "function" instead of "equation" is that it was easier (for me) to derive the equation from the function rather than the other way around. I agree the inconsistency is not good though. —Celtic Minstrel (talk &#x2022; contribs) 20:30, 6 December 2007 (UTC)


 * And for the record, the only one of those that doesn't have a "function" page is Quintic:
 * Linear function
 * Quadratic function
 * Quartic function
 * Quintic function is a redirect.
 * —Celtic Minstrel (talk &#x2022; contribs) 20:44, 6 December 2007 (UTC)


 * I wasn't aware of the function pages; I only came across this one when I saw the suggested merge at 'Cubic equation'. I've seen that you've proposed the merging of quartic function/equation and I think as long as we're consistent about article names, all will be fine. asyndeton   talk  20:48, 6 December 2007 (UTC)

Cardano/Tartaglia Contest
I saw no source for what was previously up, and what I'm saying has a source for it, so I believe I'm justified in correcting what was said.

What was previously said was that Cardano broke his promise to Tartaglia. By modern standards, this is incorrect. The promise he made to Tartaglia was that he would not publish Tartaglia's work: Cardano published del Ferro's work. Not only that, but part of the agreement was that even if he did publish something on cubics, that he give Tartaglia a decent amount of time to at least publish his results. And I don't believe it was just a few years, he gave Tartaglia about a decade to publish his results, in which Tartaglia did nothing. Cardano even published a book about arithmetic in the meantime and sent Tartaglia a copy to show that he kept his promise. That, and the fact that he could've ignored giving Tartaglia any credit for an independent proof should be something worthwhile to say.

Later on I'll add in a few more dates and specifics from my source. Fephisto (talk) 03:20, 4 March 2008 (UTC)


 * There is one source that does say that Tartaglia broke his promise, that is Mathematics: From the Birth of Numbers by Jan Gullberg, page 316. It is prudent to say there have been several areas in this book that have been inaccurate (still an excellent book), so I am not saying you are wrong, I am just stating that there is a published book that contradicts what you are saying.  So what is your source for the information you provided?--Terets (talk) 17:15, 28 December 2008 (UTC)


 * To be a sufficiently accurate source it needs to come from a peer-reviewable source like a scholarly journal. Published books are not bound to this requirement and thus should not be used as a primary source of information. --Fusionshrimp 128.194.39.250 (talk) 20:02, 2 March 2009 (UTC)

The (complete) root-finding formula
I've written the complete formula that treat the $$\Delta<0$$ case too. Tested and working, found on an Italian math paper. Sorry for my bad English and format. Please, make it more readable if you mind... Heavymachinegun 87.10.134.182 (talk) 23:49, 4 July 2008 (UTC)

Four methods for solution of the cubic equation by Nilaish
Method is shown by the help of the article given below :-

Abstract & Article :

We have general cubic equation ,

(1) ax³ + bx² + cx + d = 0 ; where a, b, c , d € R , a ≠ 0. Also ,

(1’) x³ + (b/a)x² + (c/a)x + d/a = 0

Solution :

Method 1

As from the equation 1, we can say that , it can factorized as shown below,

(2) δ( x + α )( x + β )( x + γ ) = 0

where δ, α, β, γ are arbitrary constants. The matter which is to be observed is given below. When we split 2 then we get after rearranging ,

(3) x³ + ( α + β + γ )x² + ( αβ + βγ + γα )x + αβγ = 0.

Hence, equations 3 and 1’ are identical so we can equate coefficients , we get,

(4) ∑α = (b/a).

(5) ∑αβ = (c/a).

(6) αβγ = (d/a).

Solving equations 4, 5, 6 which are linear equations in four variables, we will get the roots of the equation 1, i.e.,

(7) x = {-α, -β, -γ }.

Method 2

As from equation 1’ we have ,

x³ + (b/a)x² + (c/a)x + d/a = 0

As from equation 3 we have ,

x³ + ( α + β + γ )x² + ( αβ + βγ + γα )x + αβγ = 0.

Now we can mentally write the roots of the given cubic equation. Factorize constant term in the equation (d/a) ; into three factors say, p, q , r.

Such that,

(8) p + q + r = (b/a). (9) pq + qr + pr = (c/a).

This method is best illustrated using the example give below ,

F(x) = x³ + 6x² + 11x + 6 = 0

Sum of coefficients = S = 1 + 6 + 11 + 6 = 24.

Constant term in the equation = 6.

Now we have, 6 = 1.2.3 = 1.1.6

Now we will check the conditions given in 8 and 9 which set of p, q , r satisfies these conditions are the roots of the equation.

And we see that ( 1,2,3 ) satisfies these conditions so these are the roots of the equation, x³ + 6x² + 11x + 6 = 0.

And ( 1,1,6) is rejected.

Note :- In using these methods always remove the coefficient of x³. It should be always 1.

Method 3

From the equation 1’ let ,

A = (b/a) ; B = (c/a) and C = (d/a).

Hence we get,

(10) x³ + Ax² + Bx + C = 0.

Now let , (11) (x³ + Ax^2 ) = − ( Bx + C )

Now complete ( x³ + Ax^2 ) in cube ,say, ( x + p )³ then,

(12) ( x + p )³ = x³ + Ax² + A’x + p³

where, Ax² = 3px² and A’x = 3p²x

Now put 7.11 in 7.12 and we get,

(13) ( x + p )³ = − ( Bx + C ) + A’x + p³

Now in practice R.H.S of the equation 13 turns into ,

(14) ( x + p )³ = ( x + p ).

So, we put ( x + p ) = y.

So the equation becomes,

Or , y³ − y = 0 Or, y²( y − 1 ) = 0.

i.e, we get this three equations simply to solve these cubic equations,

(15) ( x + p ) = 0. (16) ( x + p ) = 1. (17) ( x + p ) = -1.

And the solution of the equation 1 is given by ,

(18) x = { -p, ( -p + 1 ) , ( -p – 1 ) }.

A few examples for illustration :-

Ex1. x³ + 6x² + 11x + 6 = 0.

ð x³ + 6x² = - ( 11x + 6 )

But, ( x + 2 )³ = x³ + 6x² + 12x + 8.

= - ( 11x + 6 ) + 12x + 8.

= ( x + 2 ).

So, y³ − y = 0. Therefore, x = { -2, -1 , 3 }.

Ex.2. x³ + 7x² + 14x + 8 = 0

Proceeding in the same way, ( x + 3 )³ = x³ + 9x² + 27x + 27.

= 2x² + 13x + 19.

= ( x + 3 )( 2x + 7 ) – 2. so, y³ = y( 2y + 1 ) – 2. or, y³ - y( 2y + 1 ) + 1 = 0. Or, ( y – 1 )( y + 1 )( y – 2 ) = 0

So, x = { -2, 4 , ± 1 }.

Note : 2( x + 3 ) – 6 + 7 = (2y + 1)

Method 4 :

Standard form of cubic equation is given by ,

(1.1) ax³ + bx² + cx + d = 0 ; Where a, b , c , d € R and a ≠ 0.

We can use many methods to solve this equation (1.1) but according to Numerical Analysis method, we require two boundaries to obtain roots.

We have from equation (1.1) ,

(1.1) ax³ + bx² + cx + d = 0 ; a ≠ 0.

And a, b , c , d € R.

Or, ax³ + bx² = - ( cx + d ).

So, x = √[ - ( cx + d )/(ax + b) ].

Or, x = √(-M). , where M = [( cx + d )/(ax + b)].

M must be negative or zero to have real outcome. Otherwise it will be some complex number.

(1.2) M ≤ 0

Which means ,

[(cx + d)/(ax + b)] ≤ 0.

Or, ( cx + d )( ax + b ) ≤ 0.

Iff, ( ax + b) ≠ 0. Now we have ,

( x + d/c )( x + b/a ) ≤ 0. c,a ≠ 0

i.e.,

(1.3) ( x + d/c )( x + b/a ) ≤ 0

we have a beautiful inequality, which can be further solved by using any iterative method. By using sign scheme we can extract the boundaries for the equation 1.1.

Note :- The only thing which must be taken care of is ( ax + b) ≠ 0 and c,a ≠ 0, > 0.

After getting the boundaries for the equation 1.3 we can further proceed using Iterative Method.

One Example to illustrate my method :-

Solve for x in, x³ - 6x² + 11x – 6 = 0.

Solution :

Using my analysis ; x³ - 6x² = -( 11x – 6 ).

So, x = √[-(11x – 6)/(x – 6 )].

As per using my equation 1.2 we have now,

Or, ( 11x – 6 )( x – 6 ) ≤ 0

Taking the sign scheme of this inequation, The solution for the inequality we have ,

So, x € [ 6/11, 6 ].

Now testing f(a) => f ( 6/11 ) <> 0.

Therefore, according to Newton - Raphson’s method at x◦ = 6. we will have a calculation chart which shows the possible iterations using the equation given below. ( 1.4 ) x1 = x◦ - f(x◦)/f’(x◦)

After seven Iterations we have, Approximate value of one root is to three decimal places is 3.000. Therefore, we get ( x – 3 ) a factor of x³ - 6x² + 11x – 6 = 0. Hence, by using synthetic Division method, we can further calculate the other roots.

Source : Dr. Nilaish's Journal for maths., ISSN 0974 - 3022, http://nilaish.livejournal.com/ , Vol.1, No. 8,7. nilaish (talk) 05:18, 4 August 2008 (UTC)

Observation
My friend from above, there is only one and only one way to solve a cubic equation, algebraically, which was given by italian's mathematicians, in then XV century, "those men's where the choseen ones, which in their hard lives had the pleasure to solve it, as a gift of destiny" I work hard to find an altenative solution, and in fact I came with one, which I do not publish, however Tartaglia condition's, are unavoidable, so is the same so this seem to me like the only way, otherwise there are, numerical methods, to do it, but !!JUST AN ALGEBRAIC GENERAL SOLUTION!!, I try what you have said but always find me, in the same cubic, you can not solve a cubic equation, nor any equation solving something in a minor degree.

Lagrange solvents, my oh, my....

Gaddy Alcalá —Preceding unsigned comment added by 200.90.51.187 (talk) 17:46, 2 May 2009 (UTC)

Correct equal case
In the "Root-finding formula" section the solution is split for DELTA >= 0 and DELTA < 0 but at the beginning of the article there's a separate case for DELTA = 0. I think this last case should be treated apart or at least put in the DELTA<0 case, where we have 3 real solutions. For instance when we have y = x^3, DELTA=0 and we get 3 coincident solutions that are real. Is it just that with DELTA = 0 in the second case we may get a NaN calculating theta ? Wentu (talk) 15:09, 22 August 2008 (UTC)

Derivative
I think that formula for x is totally wrong. Should be $$x=\frac{-2b \pm \sqrt {4b^2-12ac\ }}{6a}$$.. Correct me, if I'm wrong.. --BiH (talk) 22:39, 29 December 2008 (UTC)
 * Pull a factor 2 out of the square root and divide numerator and denominator by 2. DVdm (talk) 10:16, 30 December 2008 (UTC)
 * Indeed.. Wasn't paying attention on that.. Thanks! --BiH (talk) 19:45, 31 December 2008 (UTC)

How to use these things
I want to use this useful piece of mathematical technology to make some graphs. Sadly, this page lacks the critical information. Could a mathematician please add a section to explain whether one changes a, b or c to make the curves flatter/steeper, change the y intercept, etc.? The article on quadratic equations does this graphically through a clever image, but a few lines of text would be great. Apologies to all mathematicians recoiling in horror at the thought of using maths.... :-) --Matt's talk 18:50, 8 January 2009 (UTC)


 * Could you explain what makes it critical information? I agree that applied math has its importance; I just don't see what makes this particular use of a graph useful. Also, the reason that such a thing is not explored for cubics is it would be considerably more complicated than quadratics as there are four degrees of freedom instead of just three. This can be easily demonstrated when one considers that a quadratic can be written easily in the form a(x-h)^2+k so that each translation or stretch factor is isolated; to do this for cubics would be horrendously ugly. --Fusionshrimp 128.194.39.250 (talk) 20:11, 2 March 2009 (UTC)


 * Although I should say that this and related articles are pretty much horrendously ugly and useless plethoras of information. —Preceding unsigned comment added by 128.194.39.250 (talk) 20:16, 2 March 2009 (UTC)

One way to do this is has been shown by Thomas Mueller at "http://demonstrations.wolfram.com/ParametersForPlottingACubicPolynomial" perhaps we should add this to the reference list? Halothane (talk) 11:31, 1 July 2009 (UTC)

question about u=0
The article states that the roots x can be found from x=-p/3u + u - a/3

What if u==0??


 * ==> Since u is defined as satisfying the condition 3uv + p = 0, your what-if u = 0 demands that p = 0, which reduces equation (2) to t3 + q = 0. DVdm (talk) 17:29, 22 January 2009 (UTC)

For example: 3x^3+10x^2+14x+27=0 and normalised to x^3+3.33x^2+4.66x+3=0


 * ==> Overhere (and with this reduced precision) this would normalise to x3 + 3.33 x2 + 4.6 7 x + 9 = 0, resulting in u = 0.173. DVdm (talk) 17:29, 22 January 2009 (UTC)

gives u=0, which results in a division by zero?


 * ==> Even your version x3 + 3.33 x2 + 4.66 x + 3 = 0 does not result in u = 0. Overhere it results in u = 0.377. DVdm (talk) 17:29, 22 January 2009 (UTC)

However, this cubic equation has 3 non-zero roots.

Can anyone provide a C++/Java implementation based on this page that actually works?

I've spent the past 2 days coding up both C++ and Java methods using several different variants based on this article and can't get any of them to give the correct answers in all test cases. —Preceding unsigned comment added by 212.20.240.70 (talk • contribs)


 * ==> I suspect you either have defined your variable u as an integer, or you have used integer division. Anyway, always use a good old calculator before you try a programming language you don't thoroughly understand. DVdm (talk) 17:29, 22 January 2009 (UTC)

Why I don't have a wikipedia account
Yesterday I added a discussion comment as to an external reference in fact discussing the solution of quartics and not cubics and the confusing statement regarding D, and today I see that my comment is deleted and the text body remains unaltered.

What is the point of having a wikipedia "by the people, for the people" if some moderator deletes text without addressing the points made?

Wikipedia is a 21st century equivalent of the dark ages scriptures transcribed by moderating monks. Free - but moderated.

Now go and moderate my free expression. —Preceding unsigned comment added by 212.20.240.70 (talk • contribs)


 * ==> Consider it done :-) - DVdm (talk) 19:10, 22 January 2009 (UTC)

This is crazy. The use of the constants is so inconsistent that an average reader would not be able to follow the derivations near the bottom of the page. The discriminant case, for example, is wrong. —Preceding unsigned comment added by 76.75.119.183 (talk) 20:36, 3 February 2009 (UTC)

Inconsistent Discriminant Definition
There is one in "Nature of the Roots" and another in "Root-Finding Formula". I think they differ by a multiplicative factor of -1/(108 a^4). A consequence is that in one case, there are 3 real roots if it is positive, but in the other case there are 3 real roots if it is negative. Shouldn't they use the same definition of the term within the same article?John Lawrence (talk) 17:45, 13 April 2009 (UTC)

Monic formula of roots error?
The first root given under the monic formula does not appear to be a solution to the monic equation. I have not tested the other two root formulas.

128.135.226.20 (talk) 22:10, 5 August 2009 (UTC)


 * Looks ok to me. Can you specify in detail how you tested this? DVdm (talk) 12:34, 7 August 2009 (UTC)


 * I took the formula as written, substituted into the monic cubic function as written using Mathematica and did not get zero.


 * Specifically, I considered:
 * $$\begin{align}

x_1 = &-\frac{a}{3}\\ &-\frac{1}{3} \sqrt[3]{\frac{2 a^3-9 a b+27 c+\sqrt{\left(2 a^3-9 a b +27 c\right)^2-4 \left(a^2-3 b\right)^3}}{2}}\\ &-\frac{1}{3} \sqrt[3]{\frac{2 a^3-9 a b+27 c-\sqrt{\left(2 a^3-9 a b +27 c\right)^2-4 \left(a^2-3 b\right)^3}}{2}}\\ \end{align}$$
 * and I defined $$\alpha$$ to be the first cube root and $$\beta$$ to be the second cube root. I then evaluated $$x^3 + a x^2 + b x + c$$ for $$x = -\frac{a}{3}-\frac{\alpha}{3}-\frac{\beta}{3}$$, did a FullSimplify command, and didn't get zero.  I have now checked this computation for a third time, still not getting zero.  The only possibility I can think of that might justify this formula being true and Mathematica not evaluating my calculations to zero is if the principal roots are defined differently, but I would have expected that to result in a permutation of the three formulas for $$x$$. 128.135.226.20 (talk) 19:39, 10 August 2009 (UTC)


 * Right. Mathematica... Maple...
 * Maple (12, 13) can't simplify to zero either. But use pen and paper and check it manually. It works perfectly.
 * With numerics it does simplify to zero in Maple (and probably in Mathematica) provided you avoid the principal roots pitfall by expressing $$\beta$$ in terms of $$\alpha$$ and the coefficients (compare with section Cubic function, where $$x=-\frac{p}{3u}+u-{a\over 3}$$, and where u plays the role of your $$-\frac{\alpha}{3}$$, and $$-\frac{p}{3u}$$ of your $$-\frac{\beta}{3}$$ ), thus making sure that you use the same root expression (u, or your alpha) in both root terms.
 * Another way to force (at least numeric) correct simplification to zero in Maple is by not using root(X,3), but surd(X,3), which gives the root closest to X, as opposed to the principal root. For instance, calculate root(-8.,3) in Maple (or (-8.)^(1/3) in Mathematica). You don't get -2. In Maple the expression surd(-8.,3) does produce -2. I don't know the corresponding function in Mathematica.
 * Enjoy, DVdm (talk) 21:17, 10 August 2009 (UTC)

Proper source for Standard and Monic?
The current source for the section General formula of roots (and hence the monic formula) is this and it points to this. Although these equations are perfectly correct, we can hardly call these sources proper. I have been looking for a reference, but couldn't find any reproducing this exact form. Does anyone have a good reference to a book? DVdm (talk) 13:47, 10 August 2009 (UTC)

Attention Step aka 77.238.*.*
Hello, Stap aka 77.238.*.*, I have removed your contributions from this talk page. Can you please:
 * have a close look at Wikipedia's talk page guidelines?
 * sign in with your username Stap before you make edits?
 * sign your comments on talk pages with 4 tildes ( ~ ) at the end of your messages?
 * stop using this article talk page as a notepad unless you have a specific proposal to modify or add something to the article, that is properly sourced and that is no original research by yourself?

Thank you. - DVdm (talk) 07:09, 21 October 2009 (UTC)

= Stap answers to DVdm ==

Are you DVdm an administrator? If so, I assumed you to take care of the significance and correctness from mathematical point of view rather than to supervise how it is signed in and/or to count how many edits one needs to improve his contribution.

As far as I know the work so carelessly removed is original research of myself and I am confident in its correctness but disappointed that nobody recognized its advantages in comparison to the article:

....Removed WP:Original research and email....

Stap77.238.214.167 (talk) 08:49, 10 November 2009 (UTC)


 * Original research is not allowed on wikipedia. The removal of your contribution was correct. This talk page is for discussion about improving the article, it is not a personal notepad. I have removed the original research and email above. Whether the original research i correct or has advantages is not the point. Anything in wikipedia must be based on WP:Reliable sources which means journals, books reliable newspapers or suchlike. Get your contribution into a jouyrnal first before trying to stick it into wikipedia. Dmcq (talk) 12:03, 10 November 2009 (UTC)

Four weeks passed by but I did not receive email as said – therefore I must replay here. Nevertheless, meanwhile I kept searching that resulted with:

Weisstein, Eric W. "Cubic Formula." From MathWorld--A Wolfram Web Resource. that is, I hope, WP:Reliable sources and can be added at the External links of the article instead or above ˝Cubic Equation˝ of same author. Article contains 93 formula-lines referring to:

Dickson, L. E. ˝A New solution of the Cubic Equation.˝ Amer. Math Monthly 5, 38.39, 1898 or Elementary Theory of Equations. New York: Wiley, pp. 36-37, 1914 as well as to

Birkhoff, G. and Mac Lane, S. A Survey of Modern Algebra, 5th ed. New York: Macmillan, pp. 90-91, 106-107, and 414-417, 1996.

Due to inability of cube rooting of complex numbers the roots of equation type where sign(b^2 – 3ac) = +1 > |9abc – 2b^3 – 27a^2d |/2| b^2 – 3ac |^1.5 > 0 can’t be obtained by means of any of the formulae given at WP article that is therefore uncompleted. In order to overcome same problem L.E. Dickson almost a century ago commencing from Depressed cubic (15) and setting substitution (75) established regular Radical form of cubic (78): 4y^3 + 3sgn(p) = (3/|p|)^1.5*q/2 = C (as a matter of fact a correction of Chebyshev one z^3 – 3z = q*p^1.5) along with roots for sgn(p) = +1 at (80): y = sinh[⅓sinh^(–1) C] and for sgn(p) = –1 at (83): y = cosh[⅓cosh^(–1) C] for C ≥ 1 y = –cosh[⅓cosh^(–1) |C|] for C ≤ –1 y = cos[⅓cos^(–1) C] [three solutions] for |C| <1. Radical form (78) is identical to 4X^3 – 3PX = (q^2/p^3)^(1/2) = T ≥ 0 in my notation where, due to unary operators Q = 0; ±1 and P = sign(b^2 – 3ac) = ±1, coefficients at (78) and sign ambiguity at (83) are avoided and +k*120°is added after C at last line. Both variables are given in terms of one determinant and its sub-determinant facilitating remembrance: 2Q*q=\begin{vmatrix}0&3a&2b\\3a&b&c\\b&c&3d\end{vmatrix}= 9abc-27a^2d-2b^3, P*p=\begin{vmatrix}b&3a\\c&b\end{vmatrix}=b^2-3ac The roots of original equation are x_k = [Q*2p^(1/2)X_k – b]/(3a) corresponding to (84): x_i = 2(|p|/3)^(1/2)*y_i – a_2/3. Another inadequacy at WP article is a choice of the functions for both graphs being, due to Q*q = 0 = T, marginal case (becoming quadratic 4X^2 – 3 = 0) that as such may not be promoted as representative ones. One of them could be replaced with a graph designed by means of Graphic explorer for cubic functions or "Cubic Equation" authorized by E.W. Weisstein too. That is, in my opinion, reason more to ask author’s permission for WP article to be completed on the basis of his works. In such case somehow could be included the determinant(s) and Bisection method of graphical resolving as per my Talk: Roots of 4X^3 = ±3X + T are equal to X-coordinates of the points where 4Y = ±3X + T cuts Y = X^3. Shifting y-axis right for b/(3a) and sliding 4Y = ±3X up for T/4 at single graph can be presented all of 6 equation types: 4x^3 + 18x^2 – 81x = 148.5; 220; 364.5; 545 where x_0 = –1.5; 4; 4.5; 5 and 4x^3 + 18x^2 + 135x = –175.5; 1084 where x_0 = –1.5; 4 or in Radical form: 4X^3 – 3X = 0; 143/432; 1; 793/432 where X_0 = 0; 11/12; 1; 13/12 and 4X^3 + 3X = 0; 2519/432 where X_0 = 0; 11/12 There are one intermediate case (T = 1 = P) but two marginal ones (T = 0). Hoping my stubbornness would be considered as benevolent but not reasonless. Mladen Stambuk, Stap77.238.216.76 (talk) 23:43, 6 December 2009 (UTC)


 * As was said before Original research is not welcome on Wikipedia. It must be published and peer reviewed first. We are not qualified to peer review your work. Dmcq (talk) 00:42, 7 December 2009 (UTC)


 * I've had another looka t what you've said above. The links look okay. Are you saying that what you ave written this time is well sourced rather than your own work? That you said what you put in before was your own original research was the reason it was removed but well cited work rephrasing in your own words work that is in books is a reasonable sort of thing to put in. Dmcq (talk) 07:59, 7 December 2009 (UTC)

I'm glad that “you've another look”. A month ago I said “as far as I know it is my original work” but now it is clear to me that L.E. Dickson revealed essentially same formulae at 1914. In algebra the truth is must: formula either does or doesn’t work but ways to get same one can be different. In this case the truth is that almost fifty years ago, being not aware of Dickson’s or Chebyshev’s even Tartaglia’s work, I passingly – in order to get a solution of integral type dx/P3(x) applying Euler’s formulae for triple argument – came to the roots in this notation that I still consider as most suitable. Assembling the variables into the determinants could be considered as original research from the point of view of an administrator but in my opinion that is well done homework only being very useful as mnemonic for the students.

I don’t care either E.W. Weisstein or main notation will be chosen - the point is that WP article should be reedited since it is almost worthless without explicit roots for equation type where

0 < |9abc – 2b3 – 27a2d|(b2 – 3ac)–1.5 < 2 along with both graphs where 9abc – 2b3 – 27a2d = 0 That is your turn to do as you have to do! I’m not going to bother any more. Regards Stap77.238.206.113 (talk) 07:32, 10 December 2009 (UTC)

Inflection Point
"if b2-3ac=0, then the cubic has one inflection point"

Does not every cubic have exactly one inflection point?

~JacekW —Preceding unsigned comment added by JacekW (talk • contribs) 21:44, 21 December 2009 (UTC)


 * Yes, good catch. I have modified the text to: "If b2-3ac=0, then the cubic's inflection point is the only critical point." DVdm (talk) 23:12, 21 December 2009 (UTC)

Source Nine
http://hk.knowledge.yahoo.com/question/question?qid=7007111502216, from which the general equations are derived, should be in English at the very least on the English encyclopaedia. --121.44.147.29 (talk) 07:36, 5 January 2010 (UTC)

Introduction
The introduction to this article is horribly written. A first glance at it does not even provide a remote definition of "cubic". And the definition of "polynomial" is also placed into question. It is inundated with a variety of links spinning elsewhere to other pages, seemingly comprehensible only to the mathematically oriented. Is there an easier-to-understand version of this article at all in it?71.108.26.48 (talk) 06:57, 22 February 2010 (UTC) —Preceding unsigned comment added by 71.118.39.165 (talk)


 * See this comment on your talk page. - DVdm (talk) 08:45, 25 February 2010 (UTC)


 * If you are on this page, I have already responded to your comment on my talk page. 71.108.11.25 (talk) 03:34, 26 February 2010 (UTC)

Misuse of sources
A request for comments has been filed concerning the conduct of. Jagged 85 is one of the main contributors to Wikipedia (over 67,000 edits, he's ranked 198 in the number of edits), and practically all of his edits have to do with Islamic science, technology and philosophy. This editor has persistently misused sources here over several years. This editor's contributions are always well provided with citations, but examination of these sources often reveals either a blatant misrepresentation of those sources or a selective interpretation, going beyond any reasonable interpretation of the authors' intent. I searched the page history, and found 8 edits by Jagged 85 in September/October 2008. Tobby72 (talk) 19:14, 10 June 2010 (UTC)
 * That's an old and archived RfC. The point is still valid though, and his contribs need to be doublechecked. Tobby72 (talk) 20:55, 10 June 2010 (UTC)
 * Question: why should we do the doublechecking? Why don't we just remove the contribs? If/when someone wants to have something restored, they can provide the proper sourcing. Woulnd't that be the proper way to take care of this? DVdm (talk) 21:02, 10 June 2010 (UTC)
 * I would not delete it but rather label the passages as needing the citations checked; perhaps a dubious tag pointing here or comments on each questionable reference. Google books has one of the references, so Tusi may belong ahead of Viete. A case analysis was done, so the discriminant remark may not be far off. The comments about numerical methods are interesting, but could suggest misstatements. Glrx (talk) 21:56, 10 June 2010 (UTC)
 * Yes, possibly, but see also User talk:Tobby72 - DVdm (talk) 22:06, 10 June 2010 (UTC)

In the contributions of Jagged 85 to this article, there is not only misuse of sources but also historical assertions which are clearly false: I read: "He discovered [in the 11-th century] that a cubic equation [...] cannot be solved using earlier compass and straightedge constructions". This means that he has shown that doubling the cube cannot be solved using earlier compass and straightedge constructions. Such a proof was impossible with the technology of that time, as it needs the notion of the degree of a field extension. By the way, the history section should mention the cubic equations which were studied by the ancient Greeks, namely doubling the cube and angle trisection.

Similarly Horner scheme is presented as a tool to approximate the root of an equation when it is only a tool to evaluate a polynomial.

Being unable to know what Omar Khayyám and Sharaf al-Dīn al-Tūsī have really done on the subject, I suggest the following:
 * Replacing the paragraphs concerning the two authors by a sentence asserting that they have results on this subject
 * Precise that doubling the cube is the simplest and oldest studied cubic equation
 * Add the mention that angle trisection is another old cubic equation problem, even it has been understood much later as such a cubic problem.D.Lazard (talk) 03:35, 11 June 2010 (UTC)


 * The bit which really looks problematic to me is the way the Egyptians are described right at the beginning of the history section. The text messes them up with the Greeks. I haven't any access to the source to see what could be said reasonably, The bit about Omar Khayyám and Sharaf al-Dīn al-Tūsī looks reasonable to me. Dmcq (talk) 09:34, 11 June 2010 (UTC)


 * The business of Omar Khayyámasnd the ruler and compass it should be more he conjectured or some word like that rather than that he discovered that the cubic could not be solved using ruler and compass. Dmcq (talk) 09:43, 11 June 2010 (UTC)