DoubleGrid 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(2\phi) = 2 + 2\left[ 2\cos^2(\phi) - 1 \right] = 4\cos^2(\phi) \\ &\hieruit& 2 + 2\cos(2\phi) = \left[\left\{2 + 2\cos(\phi)\right\} - 2\right]^2 \end{eqnarray*} Hyperbolic case: \begin{eqnarray*} && 2 + 2\cosh(2p) = 2 + 2\left[ 2\cosh^2(p) - 1 \right] = 4\cosh^2(p) \\ &\hieruit& 2 + 2\cosh(2p) = \left[\left\{2 + 2\cosh(p)\right\} - 2\right]^2 \end{eqnarray*} Summarizing both cases: \begin{eqnarray*} y(\phi) = 2 + 2\cos(\phi) &\hieruit& y(2\phi) = \left[\,y(\phi)-2\,\right]^2 \\ y(p) = 2 + 2\cosh(p) &\hieruit& y(2p) = \left[\,y(p)-2\,\right]^2 \end{eqnarray*} The product of the outer diagonal elements with DoubleGrid Calculus has been presented in the paragraph Product Function of the corresponding document: $$ a'.b' = \frac{a^2}{1-2.a.b}\;\frac{b^2}{1-2.a.b} = \left(\frac{a.b}{1-2.a.b}\right)^2 $$ Thus the grid doubling iterations are of the form: $$ x := \left(\frac{x}{1-2x}\right)^2 = \left(\frac{1}{1/x-2}\right)^2 $$ The latter meaning that we could consider instead the iterations: $$ 1/x := (1/x-2)^2 \OF y := (y-2)^2 $$ Now remember the result we have just derived: \begin{eqnarray*} y(\phi) = 2+2\cos(\phi) &\hieruit& y(2\phi) = \left[\,y(\phi)-2\,\right]^2 \\ y(p) = 2+2\cosh(p) &\hieruit& y(2p) = \left[\,y(p)-2\,\right]^2 \end{eqnarray*} And it is apparent 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 $y \ge 4$. Nothing new, actually. This is entirely equivalent with statements about DoubleGrid in The Trigonometric Connection and The Hyperbolic Connectionwithin the corresponding document.
It's a matter of routine now to prove that there is a set of closed formulas for the DoubleGrid iterands. Start with $y = y(\phi)$ or $y = y(p)$ as the zero'th iterand. Then we have $y(2\phi)$ or $y(2p)$ as the first iterand, $y(4\phi)$ or $y(4p)$ as the second iterand, and so on. In general: $y(2^n\phi)$ or $y(2^np)$ as the n-th iterand. Working back to the original variables, we have for the product of the outer-diagonal elements, after $n$ grid doublings: $$ x_n = \left\{ \begin{array}{ll} 1/\left[\,2 + 2\cosh(2^n p)\,\right] & \qquad \mbox{for} \quad 0 < x_n \le 1/4 \\ 1/\left[\,2 + 2\cos(2^n \phi)\,\right] & \qquad \mbox{for} \quad x_n \ge 1/4 \end{array} \right. $$ The essentials are also found in an old 'sci.math' poster titled Re: Induction with a hard start.
It is established in MultiGrid Calculus that values $x_n$ cannot "escape" from their intervals. Thus either all iterands are $0 < x_n \le 1/4$ or all iterands are $x_n \ge 1/4$. And there is no way out. Furthermore, the formulas indicate that it's more handsome to start with any (hyperbolic) angle and work from there - simply by doubling the (hyperbolic) angles - rather than trying to determine an initial (hyperbolic) angle $p$ or $\phi$ from: $\cosh(p) = 1/(2.x)-1$ or $\cos(\phi) = 1/(2.x)-1$.