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.

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.

Discrete connections (part VII)

Starting with this part, we will deal with adjacent theories developed in order to shed some light on several of the asymptotic properties of the euclidean summation. Let us consider the following result:

[;\forall (x_n)_{n\ge 1} \subseteq \mathbf{N} \; the \; following \; assertions \; are \; equivalent;]

a)

[;\exists n_0 \in \mathbf{N} \; such \; that \; \frac{n}{2}<x_n<n, \; \forall n\ge n_0 \; and \lim_{n\rightarrow \infty}\frac{x_n + x_{x_n}}{n}=1;]

b)

[;\lim_{n\rightarrow \infty} \frac{x_n}{n} = \phi;]

Let us first prove that a) implies b). Indeed, since:

[;\frac{1}{2}<\frac{x_n}{n}<1, \; \forall n\ge n_0;]

results that the sequence is bounded, hence we can extract a convergent subsequence such as:

[;(k_n)_{n\ge 1} \; strictly \; increasing \; and \; (\frac{x_{k_n}}{k_n})_{n\ge 1} \rightarrow x;]

Through some computation (long but fairly simple) results that:

[;(\frac{x_{x_{k_n}}}{x_{k_n}})_{n \ge 1} \rightarrow \frac{1-x}{x};]

But out of this sequence we can extract a canonical subsequence such as:

[;(l_n)_{n\ge 1} \; strictly \; increasing \; and \; (\frac{x_{l_n}}{l_n})_{n\ge 1} \rightarrow \frac{1-x}{x};]

Therefore we will get successively the limits:

[;x, \frac{1-x}{x}, \ldots, \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}, \ldots;]

Since all of them must fall between 1/2 and 1 we get that x is bounded successively by consecutive ratios  of Fibonacci numbers, hence x equals the golden ratio conjugate.

The other implication (b) implies a)) is straightforward.

From now on we will call a Fibonacci decomposition of n any sequence verifying either a) or b).
The set of such sequences is non void since:

[;x_n=\lfloor n \cdot \phi \rfloor;]

and

[;y_n=\lceil n \cdot \phi \rceil;]

both verify b).

On the other hand, this class of sequences provides a tool for proving the conjectured asymptotic nature of the positional peek, i.e. we are left to prove that:

[;\lim_{n \rightarrow \infty} \frac{\tau_3(n) + \tau_3(\tau_3(n))}{n} = 1;]

Next: Left and right euclidean mappings

Discrete connections (part VI)

The main block of the euclidean fractal
For any four complex numbers u,v,w and z shaping a trapezoid let us consider the entity defined by the following elements:

the two base endpoints

[; u \; and \; z;]

the two top endpoints

[;v \; and \; w;]

the top line

[;\delta(v,w);]

the parallelism master line

[;\delta(u,v);]

the top value

[;\beta \in \sigma(v,w) \; such \; that \; \delta(\alpha,\beta) \| \delta(u,v);]

where:

[;\delta(z_1,z_2) \; - \; the \; complex \; line \; defined \; by \; z_1 \; and \; z_2;]

[;\sigma(z_1,z_2) \; - \; the \; open \; complex \; line \; segment \; defined \; by \; z_1 \; and \; z_2;]

[;\alpha \; -\; the \; limit \; of \; the \; top \; value \; base \; sequence ;]

The self similarity block of rank k of the euclidean fractal

For any four complex numbers u,v,w and z shaping a trapezoid let us consider the entity defined by the following elements:

the two base endpoints

[;\tilde{\alpha}_{k+1} \; and \; \tilde{\beta}_{k+2};]

the two top endpoints

[;\tilde{\beta}_{k+1} \; and \; 2\cdot \tilde{\beta}_{k+2} - \tilde{\alpha}_{k+2};]

the top line

[;\delta(\tilde{\beta}_{k+1}, 2 \cdot \tilde{\beta}_{k+2} - \tilde{\alpha}_{k+2});]

the parallelism master line

[;\delta(\tilde{\alpha}_{k+1}, \tilde{\beta}_{k+1});]

the top value

[;\beta^{(k)} = \overline{\sigma}(\tilde{\alpha}_0,\beta) \cap \sigma(\tilde{\beta}_{k+1}, 2 \cdot \tilde{\beta}_{k+2} - \tilde{\alpha}_{k+2});]

where:

[;\beta \; - \; defined \; as \; above;]

[;\overline{\sigma}(z_1,z_2) \; the \; closed \; complex \; line \; segment \; defined \; by \; z_1 \; and \; z_2;]

Moreover, the self similarity block of rank k is by itself an euclidean fractal main block.
Now for its link to the euclidean summation, consider the euclidean fractal main block defined by u=0,v=i,w=1+i and z=1. (suggestive figure pending)
Next: Fibonacci decomposition of n