Discrete connections (part IX)

First let us extend B) to (0,1).
[;Case\;  x = \frac{1}{2};]
I this case we have:

[;\left\{\begin{array}{rcl}\varepsilon_l(\frac{1}{2}) & = & 0 \\ (1 + p \cdot \frac{1}{2}) \cdot \varepsilon_l(\frac{\frac{1}{2}}{1+p\cdot \frac{1}{2}}) & = & 0 \\ \varepsilon_r(\frac{1}{2}) & = & \frac{1}{2} \\ (1 + p \cdot \frac{1}{2}) \cdot \varepsilon_r(\frac{\frac{1}{2}}{1 + p \cdot \frac{1}{2}}) & = & \frac{1}{2}\end{array}\right.;]

hence the conclusion.

[;Case \; x \in (0, \frac{1}{2});]

For p=0 we are left to prove an identity.

In the other subcases we have:

[;x \in (0,\frac{1}{2}) \Rightarrow (1-x) \in (\frac{1}{2},1) \Rightarrow^{C)};] 

[; \left\{\begin{array}{rcl} \varepsilon_l(1-x) & = & (1 + p \cdot x) \cdot \varepsilon_l(\frac{1+(p-1)\cdot x}{1 + p \cdot x}) \\ \varepsilon_r(1-x) & = & (1+p \cdot x) \cdot \varepsilon_r(\frac{1 + (p-1) \cdot x}{1 + p \cdot x}) \end{array} \right. ;]

For p greater or equal to 1 we get:

[;\frac{x}{1 + p \cdot x} \in (0,\frac{1}{2}) \Rightarrow;]

[;\frac{1+ (p-1) \cdot x}{1 + p \cdot x} \in (\frac{1}{2},1);]

Therefore:

[;\left\{ \begin{array}{rcl}\varepsilon_l(1-x) & = & x + \varepsilon_r(x) \\ \varepsilon_r(1-x) & = & x + \varepsilon_l(x) \\ \varepsilon_l(\frac{1+(p-1) \cdot x}{1 + p \cdot x}) & = & \frac{x}{1+p\cdot x} + \varepsilon_r(\frac{x}{1 + p \cdot x}) \\ \varepsilon_r(\frac{1+(p-1) \cdot x}{1 + p \cdot x}) & = & \frac{x}{1+p \cdot x} + \varepsilon_l(\frac{x}{1+p \cdot x})\end{array}\right.;]

hence the conclusion.

Having this proved we can state a substitute result for A):

A') (outer row recurrence)
[;\forall x \in (\frac{1}{2},1), \left\{\begin{array}{rcl} \varepsilon_l(x) & = & (1-x) + x \cdot \varepsilon_r(\frac{1-x}{x}) \\ \varepsilon_r(x) & = & (1-x) + x \cdot \varepsilon_l(\frac{1-x}{x}) \end{array}\right.;]

This holds since B) for (1-x) and p=1 implies:

[;\left\{\begin{array}{rcl}\varepsilon_l(1-x) & = & x \cdot \varepsilon_l(\frac{1-x}{x}) \\ \varepsilon_r(1-x) & = & x \cdot \varepsilon_r(\frac{1-x}{x}) \end{array}\right.;]

Now we can focus on the corner stone result for the left and right mappings:

[;\varepsilon_l(x) = 1 \; or \; \varepsilon_r(x) = 1 \Rightarrow x=\phi;]

Let x be a value such that the upper bound of either the left or the right mapping is reached. This value is different than 1/2 since:

[;\left\{ \begin{array}{rcl}\varepsilon_l(\frac{1}{2}) = 0 \\ \varepsilon_r(\frac{1}{2}) = \frac{1}{2}\end{array}\right.;]

Moreover x is greater than 1/2 since otherwise we would get:

[;\varepsilon_r(1-x) & = & 1 + x > 1;]

or

[;\varepsilon_l(1-x) & = & 1 + x > 1;]

thus contradicting 0).

From A') results that:

[;\varepsilon_r(\frac{1-x}{x}) = 1;]

or

[;\varepsilon_l(\frac{1-x}{x}) = 1;]

By mathematical induction one can prove that:

[;\varepsilon_l(\frac{(-1)^{k+2}\cdot F_{k+1} + (-1)^{k+3} \cdot F_{k+2} \cdot x}{(-1)^{k+1}\cdot F_{k} + (-1)^{k+2} \cdot F_{k+1} \cdot x})=1;]
or
[;\varepsilon_r(\frac{(-1)^{k+2}\cdot F_{k+1} + (-1)^{k+3} \cdot F_{k+2} \cdot x}{(-1)^{k+1}\cdot F_{k} + (-1)^{k+2} \cdot F_{k+1} \cdot x})=1;]

Since all values of x for which the upper bound of either the left or the right mapping is reached fall between 1/2 and 1 results that x is bounded successively by ratios of consecutive elements of the Fibonacci sequence, hence the conclusion.

The principle behind the proof (bounding values to Fibonacci ratios) was already applied in part IV and part VII, thus confirming again the strong connection between the mathematical concepts described throughout the paper.

Next: Left and right mapping candidates.