Talk:Cubic function

Would like to rate this, but wonder if it should be merged with cubic equation? Geometry guy 01:07, 21 May 2007 (UTC)

What is the name of the shape in a Cubic Function Graph? 168.170.46.5 16:39, 11 April 2006 (UTC)

Historically, it is called a cubic parabola, which is confusing if the term parabola has always meant the curve of a second-degree polynomial in your education. The 1911 Britannica article Parabola describes many curves having the name parabola. Alternative names include cubical parabola, parabola of third degree, cubic polynomial curve, third-degree polynomial curve, and polynomial curve of third degree. Other names are variations of what I have listed but substituting order for degree, or rephrasing of third degree as of degree 3.Nicknicknickandnick 21:02, 23 May 2006 (UTC)

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 • 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 $y^{6}+qy^{3}-{p^{3} \over 27}=0$ 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)

Chebishew radicals - inconsistency and replacement

Support to Halotane for a, b, c, d form that should be extended to "Solution in terms of Chebyshew radicals" and to "The case of Cubic equation with real coefficients".
Since Chebyshev cosine is rather inconvenient there is a suggestion below how both could be replaced with ordinary trigonometric and hyperbolic operators.
$\,P_{3}(x)=ax^{3}+bx^{2}+cx+d=0|*3^{3}a^{2}{\mbox{ gives Depressed cubic for }}3ax+b$
$\,P_{3}(3ax+b)=(3ax+b)^{3}-3(b^{2}-3ac)(3ax+b)-(9abc-27a^{2}d-2b^{3}){\mbox{ or }}$
$P_{3}(3ax+b)=(3ax+b)^{3}-3pP(3ax+b)-2qQ=P_{p}(x)=0\quad (1){\mbox{ where }}$
$Q={\frac {|9abc-27a^{2}d-2b^{3}|}{2}},{\mbox{ }}q=\pm 1{\mbox{ or }}q=0,{\mbox{ }}P=|b^{2}-3ac|,{\mbox{ }}p=\pm 1{\mbox{ but if }}$
$\,p=0,{\mbox{ }}P_{p}(x){\mbox{ is reduced into Primitive cubic with its roots for }}k=0;1;2:$
$x_{k}={\frac {\sqrt[{3}]{2qQ}}{3a}}e^{2k\pi \over 3}-{\frac {b}{3a}}\quad (2){\mbox{ where }}e^{\pm {2k\pi \over 3}}=\operatorname {cos} {2k\pi \over 3}\pm i\operatorname {sin} {2k\pi \over 3}={\sqrt[{3}]{1}}$
$\,{\mbox{ Cardano formulae in terms applied here could be merged into:}}$
$x_{k}={\frac {{\sqrt[{3}]{qQ+{\sqrt {Q^{2}-pP^{3}}}}}*e^{+i{2k\pi \over 3}}+{\sqrt[{3}]{qQ-{\sqrt {Q^{2}-pP^{3}}}}}*e^{-i{2k\pi \over 3}}-b}{3a}}\quad (3)$
${\mbox{ Supstituting }}3ax+b=2q{\sqrt {P}}X{\mbox{ and dividing with }}2q{\sqrt {P^{3}}}{\mbox{ if P ≠ 0 we get: }}$
$4X^{3}-p*3X=T={\frac {Q}{\sqrt {P^{3}}}}={\frac {|9abc-27a^{2}d-2b^{3}|}{2{\sqrt {|b^{2}-3ac|^{3}}}}}\lesseqgtr 1{\mbox{ for }}p=\pm 1\quad (4)$
$\,{\mbox{ Identical structure of the formulae for cos3U, cosh3V and sinh3W leads up }}$
${\mbox{ to three equation types depending upon either }}p=+1\lesseqgtr T\geq 0{\mbox{ or }}p=-1.$
${\mbox{ Since }}x={\frac {2q{\sqrt {P}}}{3a}}X-{\frac {b}{3a}}{\mbox{ real roots of these equation types will be: }}$
$4X^{3}-3X=T_{U}=\operatorname {cos} {3U}\leq 1,{\mbox{ }}x_{kU}={\frac {2q{\sqrt {P_{U}}}}{3a}}\operatorname {cos} {\frac {\operatorname {arccos} {T_{U}+2k\pi }}{3}}-{\frac {b}{3a}}\quad (5)$
$4X^{3}-3X=T_{V}=\operatorname {cosh} {3V}\geq 1,{\mbox{ }}x_{0V}={\frac {2q{\sqrt {P_{V}}}}{3a}}\operatorname {cosh} {\frac {\operatorname {arccosh} {T_{V}}}{3}}-{\frac {b}{3a}}\quad (6)$ $4X^{3}+3X=T_{W}=\operatorname {sinh} {3W}\geq 0,{\mbox{ }}x_{0W}={\frac {2q{\sqrt {P_{W}}}}{3a}}\operatorname {sinh} {\frac {\operatorname {arcsinh} {T_{W}}}{3}}-{\frac {b}{3a}}\quad (7)$
$\,{\mbox{ If real root is known, the other two can be obtained as follows: }}$
${\frac {P_{p}(x)-P_{p}(x_{0})}{3a(x-x_{0})}}={\frac {(3ax+b)^{3}-(3ax_{0}+b)^{3}-3pP[(3ax+b)-(3ax_{0}+b)]}{(3ax+b)-(3ax_{0}+b)}}=$
$\,=(3ax+b)^{2}+(3ax+b)(3ax_{0}+b)+(3ax_{0}+b)^{2}-3pP=0$
$3ax_{1;2}+b={\frac {-(3ax_{0}+b)\pm {\sqrt {(3ax_{0}+b)^{2}-4(3ax_{0}+b)^{2}+4*3pP}}}{2}}{\mbox{ i.e. }}$
$3ax_{1;2}=-b-{\frac {3a}{2}}x_{0}-{\frac {b}{2}}\pm {\sqrt {3pP-{\frac {3}{4}}(3ax_{0}+b)^{2}}}{\mbox{ divided with 3a gives: }}$
$x_{1;2}=-{\frac {1}{2}}x_{0}-{\frac {b}{2a}}\pm {\sqrt {{\frac {pP}{3a^{2}}}-{\frac {3}{4}}\left(x_{0}+{\frac {b}{3a}}\right)^{2}}}{\mbox{ conjugate if }}Q^{2}>pP^{3}\quad (8)$
$\,{\mbox{ (2), (5), (6), (7) and (8) substitute merged Cardano formulae (3) as well.}}$
$\,{\mbox{ Important: Formulae (5), (6) and (7) are applicable if p ≠ 0 only. }}$
Stap (talk) 08:36, 16 September 2008 (UTC)

The formula

If we have

$ax^{3}+bx^{2}+cx+d=a(x-x_{1})(x-x_{2})(x-x_{3})=0,\,$

let

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

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

$x_{1}=s+t-{\frac {b}{3a}},$

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

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

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 • 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 • 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)

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)

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)

Citations

I notice a conspicuous lack of citations throughout this page. If anyone knows where published forms of what is printed in this article can be found and is willing to go though and create citations, that would greatly improve the article as a whole.