Trigonometric Connection I

$ \def \half {\frac{1}{2}} \def \kwart {\frac{1}{4}} \def \EN {\quad \mbox{and} \quad} \def \OF {\quad \mbox{or} \quad} \def \slechts {\quad \Longleftrightarrow \quad} $ A bit of elementary trigonometry is needed in order to acquire further knowledge. Start with: $$ \cos(2.\phi) = 2.\cos^2(\phi) - 1 $$ Solve for $\cos(\phi)$, divide $\phi$ by two and take care of the signs: $$ \cos(\half \phi) = \sqrt{ \half + \half \cos(\phi)} \quad \mbox{for} \quad 0 \le \phi \le \pi $$ Augmented with: $$ \cos(\pi - \phi) = - \cos(\phi) $$ But suppose we are rather interested in the function $D(h)=2+2.\cos(\pi.h)$, restricted to the range $0 \le h \le 1$. Rewrite the above formula as such: $$ 2 + 2.\cos(\pi.\half h) = 2 + \sqrt{ 2 + 2.\cos(\pi.h)} \quad \mbox{for} \quad 0 \le h \le 1 $$ It follows that: $$ D(\half h) = 2 + \sqrt{D(h)} \quad \mbox{for} \quad 0 \le h \le 1 $$ Augmented with: $$ 2 + 2.\cos\left[\pi(1 - h) \right] = 2 - 2.\cos(\pi.h) \quad \OF \quad D(1 - h) = 4 - D(h) $$ The latter formula can also be implemented in a more "symmetric" way, as opposed to $D(h/2) = 2 + \sqrt{D(h)}$ : $$ D(1 - \half h) = 4 - D(\half h) = 2 - \sqrt{D(h)} \quad \mbox{for} \quad 0 \le h \le 1 $$ Elementary values are: $$ D(0) = 4 \qquad D(\half) = 2 \qquad D(1) = 0 $$ Start with $h = 1/2$ then: $$ D(\kwart) = 2 + \sqrt{D(\half)} = 2 + \sqrt{2} \qquad D(\frac{3}{4}) = 4 - D(\kwart) = 2 - \sqrt{2} $$ Let's try now for fractions $\times 1/8$. The values D(0/8), $D(2/8)$, D(4/8), D(6/8) and D(8/8) have already been calculated. Go for the rest: $$ D(\frac{1}{8}) = 2 + \sqrt{D(\kwart)} = 2 + \sqrt{2 + \sqrt{2}} \qquad D(\frac{7}{8}) = 4 - \sqrt{D(\frac{1}{8}} = 2 - \sqrt{2 + \sqrt{2}} $$ $$ D(\frac{3}{8}) = 2 + \sqrt{D(\frac{3}{4})} = 2 + \sqrt{2 - \sqrt{2}} \qquad D(\frac{5}{8}) = 4 - D(\frac{3}{8}) = 2 - \sqrt{2 - \sqrt{2}} $$ The outcomes can be sorted, in descending order, or with the $h$-values as a key: \begin{eqnarray*} D(0/8) &=& 4 \\ D(1/8) &=& 2 + \sqrt{2 + \sqrt{2}} \\ D(2/8) &=& 2 + \sqrt{2} \\ D(3/8) &=& 2 + \sqrt{2 - \sqrt{2}} \\ D(4/8) &=& 2 \\ D(5/8) &=& 2 - \sqrt{2 - \sqrt{2}} \\ D(6/8) &=& 2 - \sqrt{2} \\ D(7/8) &=& 2 - \sqrt{2 + \sqrt{2}} \\ D(8/8) &=& 0 \end{eqnarray*} It is evident that such a procedure will work for any denominator of the form $2^N$ where $N >0$ is an integer. The proof is by complete induction. Suppose that the work has already been done for $2^{N-1}$, then all function values for angles $\pi.2.k/2^N$ are known. The values for $\pi.k/2^N$ can be calculated by using $D(k/2^N)=2+\sqrt{D(2.k/2^N)}$, giving all values $k/2^N$ in the interval $0 The structures of square-roots-of-two are very much alike, except for the $\pm$ signs. Hence they may be abbreviated as follows and interpreted as binary codes or "numbers": $$ 2 + \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{ 2 - ... }}}} \quad \equiv \quad + - + + - \: ... \quad \equiv \quad 0 \, 1 \, 0 \, 0 \, 1 \: ... $$ Calculations for denominators of higher degree can be automated by a computer program. By doing so, it is observed that the "new" numbers, with $k = $ odd, together form a binary-reflected Gray-code. A Gray-code is characterized by the property that only one "bit" is changed at a time, while going to the next codeword (provided that both codewords are of equal length). Suppose we have a $N$-bit Gray code which is represented by a $(N \times 2^N)$ matrix of '$+$'s and '$-$'s, in such a way that the $i$'th column is the $i$'th codeword: $$ G(N) = \left[ \begin{array}{cccccc} G_0 & G_1 & ... & G_i & ... & G_{2^N-1} \end{array} \right] $$ The order of the codes belonging to $G(N)$ remains unchanged while using the formula $D(h/2)=2+\sqrt{D(h)}$, the order is reversed while using the formula $D(1-h/2)=2-\sqrt{D(h)}$, for obtaining the next generation of codewords. Thus, while performing the next step of the angle-refinement procedure, the code is augmented, for odd indices $k$, with '$+$' and '$-$' signs as follows: $$ G(N+1) = \left[ \begin{array}{ccccccccc} G_0 & G_1 & ... & G_{2^N-1} & G_{2^N-1} & G_{2^N-2} & ... & G_1 & G_0 \\ + & + & ... & + & - & - & ... & - & - \end{array} \right] $$ If $G(N)$ is a $N$-bit Gray code, then herewith it is clear that $G(N+1)$ must be a $(N+1)$-bit Gray code. Our proof by induction is completed by verifying the existence of a trivial $1$-bit code: $$ G(1) = \left[ \begin{array}{cc} + & - \end{array} \right] $$ Let's consider again now the functions which relate the the product of the off-diagonal coefficients at the coarser grid to the product of coefficients at the finer grid and vice versa. Start with: $$ x = \frac{ \sqrt{x'} }{ 2.\sqrt{x'} \pm 1} = \frac{1}{2 \pm 1/\sqrt{x'} } $$ The accompanying process of grid refinement may also be described as follows: $$ \frac{1}{x} = 2 \pm \sqrt{\frac{1}{x'}} $$ Start with $x'=1/2$ or $1/x'=2$, the value for which the (inverse) coarsening process explodes, then $1/x = 2 \pm \sqrt{2}$. Iterating again gives: $1/x := 2 \pm \sqrt{2 \pm \sqrt{2}}$. And so on and so forth. If these values are sorted in descending order, and plotted on a computer screen, then the resemblance with a cosine function readily becomes apparent. Indeed, the sequence: $$ 2 \pm \sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \sqrt{ 2 ... }}}}} $$ is recognized to be quite the same one as has been produced by iterations with the function $D$: start with $h = 1/2$ and do exactly the same with: $$ D(\half h) = 2 + \sqrt{D(h)} \OF D(1 - \half h) = 2 - \sqrt{D(h)} $$ We may conclude that there exists an intimate, one-to-one relationship between two seemingly quite different functions. The first one maps the catastrophic value $1/2$ onto points in the space of products of off-diagonal coefficients, which may then called "dangerous". The second one calculates all successive values of $2 + 2.\cos(k.\pi/2^N)$ for all $k$ with $0 \le k \le 2^N$ and $N$ an arbitrary large integer, starting with the outcomes $2 \pm \sqrt{2}$. In fact, nothing prevents us from writing: $$ \frac{1}{a'.b'} \equiv D(h) \EN \frac{1}{a.b} \equiv D(\half h) $$ Where $a.b$ denotes the off-diagonal coefficients product, which is obtained by refinement of the grid belonging to $a'.b'$. It is observed that grid-refinement corresponds with halving an angle. It may be questioned if the reverse is also true: does grid coarsening correspond with doubling an angle? $$ x' = \left( \frac{x}{2.x-1} \right)^2 \slechts \frac{1}{x'} = \left(2 - \frac{1}{x} \right)^2 $$ Which is equivalent with: $$ \cos(2.\phi) = 2.\cos^2(\phi) - 1 \slechts 2 + 2.\cos(2.\phi) = 2 + 4.\cos^2(\phi) - 2 = $$ $$ \left[ 2.\cos(\phi) \right]^2 = \left( 2 - \left[ 2 + 2.\cos(\phi) \right] \right)^2 \slechts D(2.h) = \left[ 2 - D(h) \right]^2 $$ Where $\phi = \pi.h$. Because $1/x \equiv h$ and $1/x' \equiv 2.h$, there is enough solid ground now to identify: $$ \frac{1}{a.b} = D(h) = 2 + 2.\cos(\pi.h) $$ Herewith we find the distribution of dangerous points in $(a.b)$-space, belonging to a grid refinement of order $N$, as has been verified by numerical experiments too: $$ a.b = \frac{1}{2 + 2.\cos(k.\pi/2^N)} \quad \mbox{with} \quad k = 0,1,2, ... , 2^N $$