TripleGrid Product

$ \def \hieruit {\quad \Longrightarrow \quad} \def \OF {\quad \mbox{or} \quad} $ With well known formulas for the trigonometric and hyperbolic functions, the following results can be derived.
Trigonometric case: \begin{eqnarray*} && 2 + 2\cos(3\phi) = 2 + 2\left[ \cos(2\phi) \cos(\phi) - \sin(2\phi) \sin(\phi) \right] = \\ && 2 + 2\left[ \left\{ 2\cos^2(\phi) - 1 \right\} \cos(\phi) - \left\{ 2 \sin(\phi) \cos(\phi) \right\} \sin(\phi) \right] = \\ && 2 + 2\left[ 2\cos^3(\phi) - \cos(\phi) - 2 \left\{ 1 - \cos^2(\phi) \right\} \cos(\phi) \right] = \\ && 2 + 2\left[ 2\cos^3(\phi)-\cos(\phi)-2\cos(\phi)+2\cos^3(\phi) \right] = \\ && 2 + 2\left[ 4\cos^3(\phi) - 3\cos(\phi) \right] = 8\cos^3(\phi) - 6\cos(\phi) + 2 \\ &\hieruit& 2 + 2\cos(3\phi) = 8\cos^3(\phi) - 6\cos(\phi) + 2 \end{eqnarray*} Try to divide the outcome by $\left[2 + 2\cos(\phi)\right]$: \begin{eqnarray*} && \frac{8\cos^3(\phi) - 6\cos(\phi) + 2}{2\cos(\phi) + 2} = 4\cos^2(\phi) - 4\cos(\phi) + 1 = \\ && \left[2\cos(\phi) - 1\right]^2 \\ &\hieruit& 2 + 2\cos(3\phi) = \left[2+2\cos(\phi)\right]\left[\left\{2\cos(\phi)+2\right\} - 3\right]^2 \end{eqnarray*} Hyperbolic case: \begin{eqnarray*} && 2 + 2\cosh(3p) = 2 + 2\left[ \cosh(2p) \cosh(p) + \sinh(2p) \sinh(p) \right] = \\ && 2 + 2\left[ \left\{ 2\cosh^2(p) - 1 \right\} \cosh(p) + \left\{ 2 \sinh(p) \cosh(p) \right\} \sinh(p) \right] = \\ && 2 + 2\left[ 2\cosh^3(p) - \cosh(p) + 2 \left\{ \cosh^2(p) - 1 \right\} \cosh(p) \right] = \\ && 2 + 2\left[ 2\cosh^3(p)-\cosh(p)-2\cosh(p)+2\cosh^3(p) \right] = \\ && 2 + 2\left[ 4\cosh^3(p) - 3\cosh(p) \right] = 8\cosh^3(p) - 6\cosh(p) + 2 \\ &\hieruit& 2 + 2\cosh(3p) = 8\cosh^3(p) - 6\cosh(p) + 2 \end{eqnarray*} Try to divide the outcome by $\left[2 + 2\cosh(p)\right]$: \begin{eqnarray*} && \frac{8\cosh^3(p) - 6\cosh(p) + 2}{2\cosh(p) + 2} = 4\cosh^2(p) - 4\cosh(p) + 1 = \\ && \left[2\cosh(p) - 1\right]^2 \\ &\hieruit& 2 + 2\cosh(3p) = \left[2+2\cosh(p)\right]\left[\left\{2\cosh(p)+2\right\} - 3\right]^2 \end{eqnarray*} Summarizing both cases: \begin{eqnarray*} y(\phi) = 2 + 2\cos(\phi) &\hieruit& y(3\phi) = y(\phi) \left[\,y(\phi) - 3\,\right]^2 \\ y(p) = 2 + 2\cosh(p) &\hieruit& y(3p) = y(p) \left[\,y(p) - 3\,\right]^2 \end{eqnarray*} The product of the outer diagonal elements with TripleGrid Calculus is derived easily with help of the Elimination paragraph: $$ a'.b' = \frac{a^3}{1-3.a.b}\;\frac{b^3}{1-3.a.b} = a.b \left(\frac{a.b}{1-3.a.b}\right)^2 $$ Thus the grid tripling iterations are of the form: $$ x := x \left(\frac{x}{1-3x}\right)^2 = x \left(\frac{1}{1/x-3}\right)^2 $$ The latter meaning that we could consider instead the iterations: $$ 1/x := 1/x (1/x-3)^2 \OF y := y (y-3)^2 $$ Now remember the result we have just derived: \begin{eqnarray*} y(\phi) = 2 + 2\cos(\phi) &\hieruit& y(3\phi) = y(\phi) \left[\,y(\phi) - 3\,\right]^2 \\ y(p) = 2 + 2\cosh(p) &\hieruit& y(3p) = y(p) \left[\,y(p) - 3\,\right]^2 \end{eqnarray*} And it is apparent again that these formulas both seem to cover the iterations. The trigonometric formula is valid for $0 \le y \le 4$. The hyperbolic formula is valid for $y \ge 4$. This must be the case, because of some results obtained in the DoubleGrid document, the paragraph Governing Equation. Here it is established that the dangerous and the safe domains of Gaussian Elimination have a meaning which is independent of grid coarsenings and refinements. It is entirely determined by the nature (i.e. discriminant) of a second order differential equation with constant coefficients, which is accompanying our uniform three-diagonal linear system of equations. It remains somewhat surprising, though, that the the domains $0 < y \le 4$ and $y \ge 4$ are independently and seamlessly recovered with TripleGrid as well.
It's a matter of routine now to prove that there is a set of closed formulas for the TripleGrid iterands. Start with $y = y(\phi)$ or $y = y(p)$ as the zero'th iterand. Then we have $y(3\phi)$ or $y(3p)$ as the first iterand, $y(9\phi)$ or $y(9p)$ as the second iterand, and so on. In general: $y(3^n\phi)$ or $y(3^n p)$ as the n-th iterand. Working back to the original variables, we have for the product of the outer-diagonal elements, after $n$ grid triplings: $$ x_n = \left\{ \begin{array}{ll} 1/\left[\,2 + 2\cosh(3^n p)\,\right] & \qquad \mbox{for} \quad 0 < x_n \le 1/4 \\ 1/\left[\,2 + 2\cos(3^n \phi)\,\right] & \qquad \mbox{for} \quad x_n \ge 1/4 \end{array} \right. $$ Again, it's more handsome to start with any (hyperbolic) angle and work from there - simply by tripling the angles - rather than trying to determine an initial angle $p$ or $\phi$ from: $\cosh(p) = 1/(2.x)-1$ or $\cos(\phi) = 1/(2.x)-1$.