Cyclic property of a cubic

Question

I was tempted to ask to algebrify, seeing Toby Mak's numeric cubic:

If $f(x)=x^3+u x+v$ , then what would $g(x)$ be in terms of $ (u,v)$ in order to be cyclic in the same way..

Answer 1

Given any root, $x_k$, of the given depressed cubic, the other two roots are given by

$$x_{k\pm 1} = \dfrac{-x_k}{2} \mp i \dfrac{\sqrt{3}}{2} \sqrt{x_k^2+\dfrac{4u}{3}}$$

The problem with this expression, is that, if you chose just the $-$ of the $\mp$ and iterated, you might initially cycle through all 3 roots, but you would quickly only oscillate between 2 of the 3 roots.

The problem appears to be root $x_2$, whose real part lies between the other two, e.g. $x_1 < x_2 < x_0$, or whose imaginary part is negative in the case of a pair of complex roots. For that root as input, one would want to use $\pm$ to keep the cycle going in the same direction:

$$x_{k\pm 1} = \dfrac{-x_k}{2} \pm i \dfrac{\sqrt{3}}{2} \sqrt{x_k^2+\dfrac{4u}{3}}$$

So a candidate $g(x)$, for a counter-clockwise cycle: $x_0 \rightarrow x_1 \rightarrow x_2 \rightarrow x_0$, would be

$$g(x) = \dfrac{-x_k}{2} - \left(-1\right)^{([k+2]\, \mathrm{mod}\, 3)} i \dfrac{\sqrt{3}}{2} \sqrt{x_k^2+\dfrac{4u}{3}}$$

Given a root, $x_k$, algorithmically determining $k$ is straight forward. Functionally determining $k$ with a compact expression is proving to be more difficult.

For what I mean by $k$ and "counter-clockwise cycle", please see this write-up:

Roots of the Cubic Equation in Terms of a Complex Hyperbolic Angle

and the accompanying Geogebra demo:

Complex Hyperbolic Angle to find Cubic Roots

Which graphically show that the roots of a cubic map to 3 points spaced $\dfrac{2\pi}{3}$ radians apart from each other on a circle.