Stationary Points

$ \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \OF {\quad \mbox{or} \quad} $ The $k$-th iterate of one divided by the product of the outer diagonal elements of our - meanwhile quite quite familiar - uniform three-diagonal matrix, is given by: $$ y_k = 2 + 2\cos(\phi_k) \OF y_k = 2 + 2\cosh(p_k) $$ Depending on whether $y$ is in the dangerous domain $0 \le y \le 4$ or in the safe domain $y \ge 4$ respectively. With DoubleGrid coarsening, the successive angles are given by doubling: $$ \phi_{k+1} = 2\phi_k \OF p_{k+1} = 2 p_k $$ Which is equivalent to the law: $$ y_{k+1} = (y_k - 2)^2 $$ With TripleGrid coarsening, the successive angles are given by tripling: $$ \phi_{k+1} = 3\phi_k \OF p_{k+1} = 3 p_k $$ Which is equivalent to the law: $$ y_{k+1} = y_k (y_k - 3)^2 $$ Stationary points are easily detected with the method of the angles. This has been discussed at length for DoubleGrid in the accompanying document. So let's give here a few TripleGrid examples. Starting with: $$ y_k = y_k (y_k - 3)^2 \slechts \left[(y_k - 3)^2 - 1\right] y_k = 0 \slechts $$ $$ y_k ( y_k^2 - 6 y_k + 8) = 0 \slechts y_k (y_k - 2)(y_k - 4) = 0 $$ Solutions: $$ y_k = \left\{\; 0 \; , \; 2 \; , \; 4 \;\right\} $$ Or equivalently, and perhaps easier: $$ \cos(3\phi_k) = \cos(\phi_k) \slechts 3\phi_k = \left\{\; \phi_k \;,\; - \phi_k + 2 \pi \;,\; \phi_k + 2\pi \;\right\} \slechts $$ $$ \phi_k = \left\{ 0 \;,\; \pi/2 \;,\; \pi \;\right\} \slechts y_k = 2 + 2\cos(\phi_k) = \left\{ 4 \;,\; 2 \;,\; 0 \;\right\} $$ However, these equations may give rise to new questions. We may ask, namely, what all solutions of the following equations are: \begin{eqnarray*} y_k (y_k - 3)^2 &=& 4 \\ y_k (y_k - 3)^2 &=& 2 \\ y_k (y_k - 3)^2 &=& 0 \end{eqnarray*} The first equation is equivalent with: $$ y_k^3 - 6 y_k^2 + 9 y_k - 4 = 0 $$ We know that $y_k = 4$ is a root. Thus another dividing polynomial can be found with a long division: $$ \frac{y_k^3 - 6 y_k^2 + 9 y_k - 4}{y_k - 4} = y_k^2 - 2 y_k + 1 = (y_k - 1)^2 $$ The second equation is equivalent with: $$ y_k^3 - 6 y_k^2 + 9 y_k - 2 = 0 $$ We know that $y_k = 2$ is a root. Thus another dividing polynomial can be found with a long division: $$ \frac{y_k^3 - 6 y_k^2 + 9 y_k - 2}{y_k - 2} = y_k^2 - 4 y_k + 1 = \left[ y_k - (2-\sqrt{3})\right] \left[ y_k - (2+\sqrt{3})\right] $$ The third equation is trivial and has $y_k = 3$ as an extra solution. In total we find with triple fold grid coarsening the following stationary values: \begin{eqnarray*} 1 \hieruit 4 \hieruit &4& \\ 2-\sqrt{3} \hieruit 2 \hieruit &2& \\ 2+\sqrt{3} \hieruit 2 \hieruit &2& \\ 3 \hieruit 0 \hieruit &0& \end{eqnarray*} Another example, as has been formulated and solved with my favorite Computer Algebra System (MAPLE):

> y1 := series(y0*(y0-3)^2,y0);
                                        2     3
                       y1 := 9 y0 - 6 y0  + y0

> y2 := series(y1*(y1-3)^2,y0,10);

                      2          3          4          5         6
  y2 := 81 y0 - 540 y0  + 1386 y0  - 1782 y0  + 1287 y0  - 546 y0  +

              7        8     9
        135 y0  - 18 y0  + y0

> solve(y0 = series(y1*(y1-3)^2,y0,10),y0);

                  1/2         1/2                             1/2         1/2
                 5           5          1/2       1/2        5           5
  0, 2, 4, 3/2 + ----, 3/2 - ----, 2 + 2   , 2 - 2   , 5/2 + ----, 5/2 - ----
                  2           2                               2           2

Impressive as it may seem, the method of angles enables us to solve this whole thing entirely by hand: $$ \cos(9\phi_k) = \cos(\phi_k) \slechts 9\,\phi_k = \pm \phi_k + n.2\pi \quad \mbox{where} \quad n = 0 \,,\, 1 \,,\, 2 \,,\, 3 \,,\ 4 $$ $$ \phi_k = \left\{\: 0 \:,\: \pi/4 \:,\: \pi/2 \:,\: 3\pi/4 \:,\: \pi \:,\: \pi/5 \:,\: 2\pi/5 \:,\: 3\pi/5 \: \:,\: 4\pi/5 \:\right\} $$ The angles with $\pi/5$ in them can be determined with help of the following document: Cosine Expansions.
Here a useful formula for our purpose is found: $$ \cos(5 x) = 5 \cos(x) - 20 \cos^3(x) + 16 \cos^5(x) $$ Substitute herein $x = \pi/10$, then we have, with $z = \cos(\pi/10)$: $$ 0 = 5 z - 20 z^3 + 16 z^5 \slechts 16(z^2)^2 - 20(z^2) + 5 = 0 \hieruit $$ $$ z^2 = \frac{20 \pm \sqrt{20^2 - 4.5.16}}{32} = \frac{5 \pm \sqrt{5}}{8} $$ Because it is certain that $z \ne 0$. Furthermore it is evident that we should take the largest root. Thus: \begin{eqnarray*} \cos(\pi/5) = 2\cos^2(\pi/10) - 1 = \frac{5 + \sqrt{5}}{4} - 1 &=& \frac{+ 1 + \sqrt{5}}{4} \\ \cos(2\pi/5) = 2\cos^2(\pi/5) - 1 = 2\frac{1 + 5 + 2\sqrt{5}}{16} - 1 &=& \frac{- 1 + \sqrt{5}}{4} \\ \cos(3\pi/5) = \cos(\pi - 2\pi/5) = - \cos(2\pi/5) &=& \frac{+ 1 - \sqrt{5}}{4} \\ \cos(4\pi/5) = \cos(\pi - \pi/5) = - \cos(\pi/5) &=& \frac{- 1 - \sqrt{5}}{4} \end{eqnarray*} Herewith we finally find: $$ y_k = 2 + 2\cos(\phi_k) = $$ $$ \left\{ 4 \:,\: 2+\sqrt{2} \:,\: 2 \:,\: 2-\sqrt{2} \:,\: 0 \:,\: \frac{5+\sqrt{5}}{2} \:,\: \frac{3+\sqrt{5}}{2} \:,\: \frac{5-\sqrt{5}}{2} \:,\: \frac{3-\sqrt{5}}{2} \:\right\} $$ Which is, of course, an exact match with the MAPLE solution.
At last it can be tracked how the stationary points, as found here, transform into each other, as calculated from the formula $\;y_k(y_k-3)^2 \Longrightarrow y_{k+1} $ : \begin{eqnarray*} && 2+\sqrt{2} \hieruit 2-\sqrt{2} \hieruit 2+\sqrt{2} \\ && \frac{5+\sqrt{5}}{2} \hieruit \frac{5-\sqrt{5}}{2} \hieruit \frac{5+\sqrt{5}}{2} \\ && \frac{3+\sqrt{5}}{2} \hieruit \frac{3-\sqrt{5}}{2} \hieruit \frac{3+\sqrt{5}}{2} \\ \end{eqnarray*}