Via the Cubic

$ \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \EN {\quad \mbox{and} \quad} \def \OF {\quad \mbox{or} \quad} $ The cubic equation for grid refinement with TripleGrid Calculus is: $$ y_{k+1} (y_{k+1} - 3)^2 = y_k $$ $$ y_{k+1}^3 - 6 y_{k+1}^2 + 9 y_{k+1} - y_k = 0 $$ Which is of the form: $$ a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0 $$ The discriminant of such a cubic equation is far less well known than the one of a quadratic equation. It may nevertheless be found on the Internet at Wikipedia: Discriminant.
Where we can copy and paste the following formula: $$ D = a_1^2 a_2^2 - 4a_0 a_2^3 - 4a_1^3 a_3 + 18a_0 a_1 a_2 a_3 - 27a_0^2 a_3^2 $$ The general theory about the discriminant of a polynomial is quite interesting in itself. The abovementioned web page offers pointers to topics like i.e. the resultant and via the resultant link the Sylvester matrix of two polynomials.
Whatever. Let's apply the general discriminant of the cubic to the specific one we have at hand with TripleGrid: $$ y_{k+1}^3 - 6 y_{k+1}^2 + 9 y_{k+1} - y_k = 0 $$ With $\left[\; a_3 = 1 \quad a_2 = - 6 \quad a_1 = 9 \quad a_0 = - y_k \; \right]$ its discriminant $D$ becomes: $$ D = 9^2 (-6)^2 - 4(-y_k)(-6)^3 - 4(9)^3 1 + 18(-y_k)9(-6)1 - 27(-y_k)^21^2 $$ $$ = 3^6 2^2 - 3^6 2^2 - 2^5 3^3 y_k + 3^5 2^2 y_k - 3^3 y_k^2 = 2^2 3^3 (-8+9) y_k - 3^3 y_k^2 $$ $$ \hieruit D = 27.y_k(4-y_k) $$ It is known that the above cubic equation has three real solutions if and only if this discriminant $D$ is positive. Meaning that the following is a necessary and sufficient condition for it: $$ 0 \le y_k \le 4 $$ Thus, with TripleGrid coarsening, the Dangerous Interval becomes apparent almost immediately, from the sign of the discriminant alone.
Key reference for the rest of this paragraph is: The Geometry of the Cubic Formula. Which is based on "A new approach to solving the cubic" by R.W.D. Nickalls (not available anymore).
Recall the cubic equation which is associated with triple grid refinement: $$ y_{k+1}^3 - 6 y_{k+1}^2 + 9 y_{k+1} - y_k = 0 $$ It is of the form: $$ a x^3 + b x^2 + c x + d = 0 $$ With $\left[\quad a = 1 \quad b = -6 \quad c = 9 \quad d = -y_k \quad\right]$ . According to the above reference, we can define a quantity $x_N$ by: $$ x_N = \frac{-b}{3a} = \frac{6}{3} = 2 $$ This means that a substitution $z = y_{k+1}-2$ or $y_{k+1} = z+2$ will bring our cubic into its reduced form: $$ y_{k+1}^3 - 6 y_{k+1}^2 + 9 y_{k+1} - y_k = (z+2)^3 - 6(z+2)^2 + 9(z+2) - y_k = $$ $$ z^3 + 6 z^2 + 12 z + 8 - 6 z^2 - 24 z - 24 + 9 z + 18 - y_k = z^3 - 3 z + 2 - y_k = 0 $$ Furthermore, a quantity $\delta$ is defined by: $$ \delta^2 = \frac{b^2 - 3 a c}{9 a^2} = \frac{36 - 27}{9} = 1 $$ Let's consider the case where $0 \le y_k \le 4$. Then there are three real solutions and we have the Casus Irreducibilis for the cubic equation at hand. According to the above reference, we must use a trigonometric substitution in order to be able to find any solutions: $$ z = 2\delta \cos(\theta) = 2 \cos(\theta) $$ Upon substitution it reads: $$ 8 \cos^3(\theta) - 6 \cos(\theta) + 2 - y_k = 0 $$ In the paragraph TripleGrid Product we found the following (well, eh) formula: $$ 2 + 2\cos(3\theta) = 8\cos^3(\theta) - 6\cos(\theta) + 2 $$ Therefore we have to solve: $$ 2 + 2\cos(3\theta) = y_k \slechts y_{k+1} (y_{k+1} - 3)^2 = y_k $$ Now define: $$ y_k = 2 + 2\cos(\phi_k) \EN y_{k+1} = 2 + 2\cos(\phi_{k+1}) $$ Then we have that $\theta = \phi_{k+1}$ and: $$ y_{k+1} (y_{k+1} - 3)^2 = y_k \slechts 2 + 2\cos(3\phi_{k+1}) = 2 + 2\cos(\phi_k) \slechts $$ $$ \cos(\phi_k) = \cos(3\phi_{k+1}) $$ Compare this with: $$ \cos(\phi_k) = \cos(2\phi_{k+1}) $$ And it seems that we have landed on solid ground: the method of angles.