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 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