Trigonometric Connection II

$ \def \half {\frac{1}{2}} \def \kwart {\frac{1}{4}} \def \komma {\quad , \quad} \def \OF {\quad \mbox{or} \quad} \def \hieruit {\quad \Longrightarrow \quad} $ A number of conclusions can be drawn by properly employing the formula: $$ a.b = \frac{1}{2 + 2.\cos(k.\pi/2^N)} \quad \mbox{with} \quad k = 0,1.2, ... , 2^N $$ The function reaches a minimum $1/4$ for the maximum value of the denominator, which is $4$. Because the minimum of the denominator is zero, it also reaches to infinity, meaning that dangerous points can be found at arbitrary density, increasing with $N$, and everywhere in the region $1/4 < a.b < \infty$ . It is concluded therefore that a region which is safe everywhere can only exist in the interval $0 \le a.b \le 4$, as we have seen (but not proved) before.
However, not all points in the region $1/4 < a.b < \infty$ share the same amount of risk. It seems reasonable to conjecture that points which are very close to $1/4$ are less "dangerous" that others. It will be shown now that this is indeed true. Points close to $1/4$ correspond with small angles. Hence any coarsening of the grid will correspond with the doubling of a small angle. The closer the point is to $1/4$, the smaller the angle will be. And hence it will require quite some doubling effort, before this point comes in real danger. To be more specific. Suppose that $a.b$ almost equals $1/4$. We want to use a Taylor expansion here. Some hand-held calculus can be avoided by devising an input for our favourite Computer Algebra System (MAPLE):

series(1/(2+2*cos(x)),x);

$$ \kwart + \frac{1}{16} x^2 + \frac{1}{96} x^4 + O(x^6) $$ In our case, suppose that we have a point $1/4 + \delta$, corresponding with an angle $\pi.1/2^N$. Then: $$ \kwart + \delta \approx \kwart + \frac{1}{16} (\pi/2^N)^2 $$ Solving for the angle: $$ \pi/2^N \approx \sqrt{16.\delta} \hieruit \ln(\pi) - N . \ln(2) \approx 0.5 \: \ln(16) + 0.5 \: \ln(\delta) \hieruit $$ $$ N \approx 0.5 \: ^2\log(1/\delta) \quad + \quad ^2\log(\pi) - 2 $$ If, for example, the deviation from $1/4$ is one part per million $= 10^{-6}$, then $\delta \approx 2^{-20} \rightarrow 0.5 \: ^2\log(1/\delta) \approx 10$. This means that a grid coarsening less than $10$ times offers no real danger. Which in turn means that the finest mesh can be allowed to consist of $2^{10}\approx 1000$ points, without possibly running into big trouble.
What more can be said about the stationary points we have found in the past: $$ \frac{1}{4} \komma 1 \komma \frac{3 \pm \sqrt{5}}{2} $$ Or, solve for the angle $\phi$ in: $$ a.b = \frac{1}{2 + 2.\cos(\phi)} = \quad \kwart \komma 1 \komma \frac{3}{2} \pm \frac{\sqrt{5}}{2} $$ A useful remark being: $$ \left( \frac{3}{2} + \frac{\sqrt{5}}{2} \right) \left( \frac{3}{2} - \frac{\sqrt{5}}{2} \right) = 1 $$ Solve for the cosines first: $$ \cos(\phi) = \quad 1 \komma - \half \komma \frac{ \pm \sqrt{5} - 1}{4} $$ The first angle is just zero: $\phi = 0$ and any doubling or halving of this angle maps it upon itself. Hence $1/4$ is a single stationary point, as we have seen.
The second angle must be $120^0$. A doubling of this angle gives $\cos(240^0) = -1/2$, which is the same value as with $\phi = 120^0$. Hence $1$ is also a single stationary point, as we have seen.
The 3rd/4th angles are somewhat more complicated. The smallest of the two will correspond with the plus sign. Doubling this one gives: $$ \cos(2.\phi) = 2.\cos^2(\phi) - 1 = 2 \left( \frac{ + \sqrt{5} - 1}{4} \right)^2 - 1 = \frac{5 - 2\sqrt{5} + 1}{8} - 1 = \frac{ - \sqrt{5} - 1}{4} $$ Which is precisely the second one. Let's double it again: $$ \cos(4.\phi) = 2.\cos^2(2.\phi) - 1 = 2 \left( \frac{ - \sqrt{5} - 1}{4} \right)^2 - 1 = \frac{5 + 2\sqrt{5} + 1}{8} - 1 = \frac{ + \sqrt{5} - 1}{4} $$ Giving back the first of the two. We find that the angles are determined by: $$ \cos(4.\phi) = \cos(\phi) \hieruit 4.\phi = k.2.\pi \pm \phi \hieruit $$ $$ \phi = k \frac{2.\pi}{5} \quad \left( \OF \phi = k \frac{2.\pi}{3} \right) \quad \qquad \mbox{where} \quad k = 1,2,4,8, \: ... $$ The product-function values corresponding with the denominator $5$ are stationary with a multiplicity of two: the first point gives the second point, the second point gives back the first point, and so on and so forth.
It is questioned now whether there exist also points which have a multiplicity of three or higher. The answer is affirmative. Take for example an angle which gives the same function value again, but only after three times doubling itself: $$ \cos(8.\phi) = \cos(\phi) \hieruit 8.\phi = k.2.\pi \pm \phi \hieruit $$ $$ \phi = k \frac{2.\pi}{7} \OF \phi = k \frac{2.\pi}{9} \qquad \mbox{where} \quad k = 1,2,4,8, \: ... $$ We have found some clues for the behaviour of points in the area $(a.b) > 1/4$, corresponding with angles $k.2.\pi/D$ and denominators $D=2,3,4,5,7,8,9, ...$. A denominator like $6$ can be produced by $1/3=2/6$ and halving the latter. The same story can be told for $10$ : $1/5=2/10$. Denominators $31$ and $33$ can be produced by $2^5.\phi=k.2.\pi \pm \phi$, corresponding with multiplicities $\le 5$. Then $3/33 = 1/11$ , providing material for angles $k.2.\pi/11$.
It is thus observed that, in the "dangerous" area, there also exist infinitely many points which are not "dangerous" at all, since they will always be mapped upon each other with any further grid coarsening. But, as has been suggested earlier, this situation is similar to unstable equilibrium in physics, like a pencil balancing on its tip.
Fractions $k/2^N$ are expanded in a computer as binary numbers. For example: $$ 0.00101001010011101010110101110110 $$ But also numbers like $1/3$ are represented, in a computer, as a finite binary fraction: $$ 0.01010101010101010101010101010101 $$ Given a finite (though maybe very large) binary representation of any number in the area $1/4 < a.b < \infty$, there is no sensible way to tell if this number is predestined to be dangerous or not. This remains true, even if we ever would have a computer at our disposal with immensely large words.
This story vaguely reminds to the (in)famous controversy between Formalists (Hilbert) and Intuitionists (Brouwer). The formalist mathematicians would say that any point in our product space is either dangerous or not dangerous. But the intuitionist mathematicians would argue that the Law of the Excluded Middle is not universally valid. It cannot be decided whether a point is dangerous or not. From a realistic point of view, one might find that the intuitionists are merely right. Our conclusion would be, then, that the behaviour of the products of real world off-diagonal coefficients $(a.b)$ is, actually, unpredictable, if their values are to be chosen in the domain $1/4 < (a.b) < \infty$.
We have also seen that there exists another, "safe" domain, which fortunately is quite distinct from the dangerous area, namely $0 \le (a.b) \le 1/4$. The boundary $1/4$ of this safe domain may even be defined with a sensible pinch of salt. And it is expected that no weird mathematical problems will show up for this area of interest.