Talk:Cubic function/Archive 5

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Error in "Q" in the monic trinomial form

There is an error in the formula for "q" in the monic trinomial form.

The denominator should be 27a^2 instead of 27a^3. please someone correct this. You can see a link here:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Or you can derive your own expression (which I did too) — Preceding unsigned comment added by 140.105.70.136 (talk) 15:08, 25 May 2012 (UTC)

The formula is correct. In your link the polynomial in y is not monic, as y³ has a as coefficient. Thus you have to divide by a the equation of the link before comparing. D.Lazard (talk) 17:01, 25 May 2012 (UTC)

Geometric interpretation

You don't need perfect vision to see that the angles θ are not the same, or that the three red lines intersection points on the circle do not form an equilateral triangle. Ssscienccce (talk) 10:41, 6 September 2012 (UTC)

Good catch. I modified the figure by moving the top red line slightly clockwise to make the angles match, and to make it look more like an equilateral triangle. - DVdm (talk) 11:33, 6 September 2012 (UTC)

In the illustration of Omar Khayyám's geometric solution, the circle is drawn so that the parabola intersects it at the top, putting the vertical line through the center of the circle. This suggests a trivial solution, x=b/(2a^2). It would be better to draw the circle with a different diameter (choose a different value of b) so as not to give this false impression. — Preceding unsigned comment added by 64.95.214.18 (talk) 18:55, 19 December 2012 (UTC)

Alternative trigonometric substitution

Instead of the cosine substitution x = u cos(θ), we can also use x = u sin(θ). But we must associate the coefficients with the identity

$4\sin ^{3}(\theta )-3\sin(\theta )+\sin(3\theta )=0$

Bad English and a bigoted insistence to CONTROL a page

To whoever keeps changing my edits back to the original:

My contributions are valid and intelligent. Why do you insist on withholding valid information from the people? You do not own this page, all people are allowed to contribute meaningful information.

And if youre going to change my wording or undo my edits, the least you could do is use proper, intelligible English. — Preceding unsigned comment added by 64.134.140.53 (talk) 23:46, 17 September 2012 (UTC)

Tartaglia's work

Interesting article. Im also interested in the historical component of this article. It speaks of Tartaglia competing with Ferrari and Cardano and Fiore. But Im curious to know what Tartaglias method was, and how was it unique from Ferro's method? — Preceding unsigned comment added by 75.172.58.58 (talk) 09:23, 25 September 2012 (UTC)

The historical component is indeed interesting -- one of my math professors spoke of these "duels" as rowdy public exhibitions with money and prestige on the line. If only the general populace were educated enough to enjoy such intellectual endeavors, we might have decent reality television today. — Preceding unsigned comment added by 71.226.189.240 (talk) 02:16, 29 September 2012 (UTC)

Alternative solution to depressed published in 2010 but with a glitch

The April 2010 issue of Resonance has an alternative solution to a depressed cubic found here http://www.ias.ac.in/resonance/April2010/p347-350.pdf

It involves finding values for u,v,w, so that the cubic is reduced to (x+u)^3 = -v(x+w)^3

Just take the cube root of both sides.

Only glitch is that if in the depressed cubic

x^3 +cx + d= 0 c is -3 times q^2 and d is -2 times q^3 then you get a tautology.

Try it to solve (I have set q = 1) x^3 - 3x - 2 = 0

and you wind up with (x+1)^3 = 1(x+1)^3

which is of no use in finding the solution (x=2 BTW).

Otherwise it works smoothly. (There is also a typo in the artciel on p. 349 where they say v when they mean w in one equation.)--WickerGuy (talk) 20:57, 9 May 2013 (UTC)