Discrete connections (part VIII)

This part deals with a class of conjectured mappings strongly connected to the euclidean summation and its properties.
Indeed, let us define by the euclidean left and right mappings the functions having the following properties:

0) (definition and upper boundary)

[;\varepsilon_l,\varepsilon_r:(0,1)\rightarrow [0,1];]

A) (inner row recurrence)

[;\forall x \in (\frac{1}{2},1), \left\{\begin{array}{rcl}\varepsilon_l(x) & = & (1-x) + \varepsilon_r(1-x) \\ \varepsilon_r(x) & = & (1-x) + \varepsilon_l(1-x) \end{array} \right.;]
B) (column periodicity)
[;\forall x\in (\frac{1}{2}, 1), \forall p \in \mathbf{N}, \left\{\begin{array}{rcl}\varepsilon_l(x) & = & (1+p\cdot x) \cdot \varepsilon_l(\frac{x}{1+p\cdot x})\\ \varepsilon_r(x) & = & (1 + p \cdot x) \cdot \varepsilon_r(\frac{x}{1+p \cdot x})\end{array}\right.;]
C) (diagonal periodicity)
[;\forall x\in (\frac{1}{2}, 1), \forall p \in \mathbf{N}, \left\{\begin{array}{rcl}\varepsilon_l(x) & = & ((p+1)-p\cdot x) \cdot \varepsilon_l(\frac{p - (p-1) \cdot x}{(p+1)-p\cdot x})\\ \varepsilon_r(x) & = & ((p+1) - p \cdot x) \cdot \varepsilon_r(\frac{p - (p-1) \cdot x}{(p+1)-p \cdot x})\end{array}\right.;]
D) (lower boundary)
[;\forall p \in \mathbf{N}, \; p \ge 2, \left\{ \begin{array}{rcl} \varepsilon_l(\frac{1}{p}) & = & 0 \\ \varepsilon_r(\frac{1}{p}) & = & \frac{1}{p} \end{array}\right.;]

First some basic properties of these functions:

E) (upper boundary is achieved)

[;\varepsilon_l(\phi)=\varepsilon_r(\phi)=1;]

D')
[; \forall p \in \mathbf{N}, \; p \ge 2 \left\{\begin{array}{rcl}\varepsilon_l(\frac{p-1}{p})=\frac{2}{p}\\ \varepsilon_r(\frac{p-1}{p})=\frac{1}{p}\end{array}\right.;]

For the second property the proof requires only to apply A) and D). For the first the proof takes the following steps:

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

[;\left\{\begin{array}{rcl}\varepsilon_l(\phi) & = & \varphi \cdot \varepsilon_l(1-\phi) \\ \varepsilon_r(\phi) & = & \varphi \cdot \varepsilon_r(1-\phi) \end{array}\right. \Rightarrow^{A)} \left\{ \begin{array}{rcl}\varepsilon_l(\phi) & = & \varphi \cdot (\varepsilon_r(\phi) - \phi^2) \\ \varepsilon_r(\phi) & = & \varphi \cdot (\varepsilon_l(\phi) - \phi^2)\end{array}\right. \Rightarrow;]

[;\left\{\begin{array}{rcl}(1+\varphi) \cdot \varepsilon_l(\phi) & = & (1+\varphi) \cdot \varepsilon_r(\phi) \\ \varepsilon_r(\phi) & = & \varphi \cdot (\varepsilon_l(\phi) - \phi^2) \end{array}\right. \Rightarrow ;]

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

Although expanding A) to values smaller than 1/2 is not trivial, this can be achieved for B), but after some intermediate result:

F) (unifying formula)

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

The proof follows the relative positions of x and 1/2.

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

In this case we get successively:

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

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

On the other hand from B) for x and p=1 we get:

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

hence the conclusion.
[;Case \; x=\frac{1}{2};]
Indeed one has:

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

hence the conclusion.

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

In this case we get successively:

[;x \in (0, \frac{1}{2}) \Rightarrow (1-x) \in (\frac{1}{2},1) \Rightarrow^{A)} \left\{ \begin{array}{rcl} \varepsilon_l(1-x) & = & x + \varepsilon_r(x) \\ \varepsilon_r(1-x) & = & x + \varepsilon_l(x) \end{array}\right.;]

On the other hand from C) for 1-x and p=1 we get:

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

hence the conclusion.
Next: Extra properties of the left and right mappings.