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April 18

Depressed Cubic.

---Sorry. Thanks. I read it and am stuck at 3 steps.

At this point Cardano imposed a second condition for the variables u and v (Why can he do this?):

3uv+p=0\,.

As the first parenthesis vanishes in (3), we get $u^{3}+v^{3}=-q$ and $u^{3}v^{3}=-p^{3}/27$ . Thus u^3 and v^3 are the two roots of the equation (Where did the q and z come from? What use is it to finding the roots?)

z^{2}+qz-{p^{3} \over 27}=0\,.

At this point, Cardano, who did not know complex numbers, supposed that the roots of this equation were real, that is that ${\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}>0\,$ . (How did he get from the above to this equation? -.-)

Oh. and I just found an this. I scrawled an equation on my desk a few weeks ago. It only says recurrence relation. Does anyone know what it is?

nk+1=1/n[(n-1)yk+yk^(x/(n-1))]

ThanksSomeone with a Question (talk)

— Preceding unsigned comment added by Someone with a Question (talk • contribs) 14:45, 18 April 2015 (UTC)[reply]

Actually Cubic Equation#Reduction to a depressed cubic and the following section do more or less say what it's used for. Namely a preliminary step to solving a cubic equation. It's analogous to completing the square for quadratic equations. --RDBury (talk) 15:38, 18 April 2015 (UTC)[reply]

I hope you don't mind I took the liberty of reformatting your mathy stuff. I'm unsure of your notation in the recurrence relation. Is it $n_{k+1}=1/n\left[(n-1)y_{k}+y_{k}^{(x/(n-1))}\right]$ ? —Tamfang (talk) 02:50, 19 April 2015 (UTC)[reply]

That's how I'd interpret it, but of course there's either a lot of elided subscripts, or it's nonsense. It's not one of the more famous recurrence relations or difference equations, but it could well be part of some arbitrary exercise. OP should get a nice moleskine or something so they don't need to scrawl on the desk :) SemanticMantis (talk) 16:17, 19 April 2015 (UTC)[reply]

I hope too you don't mind I took the liberty of reformatting your square brackets. ;) CiaPan (talk) 10:18, 20 April 2015 (UTC)[reply]