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 IV)

Before dealing with the major results, let us consider the following:

 Intermediate recurrence

[;\frac{n}{2}<k<\frac{2\cdot n}{3} \Rightarrow;]

[;\tau_1(n,k)=k+\tau_1(n-k,2\cdot k - n);]

Its proof consists in applying the outer row recurrence first for n and k and then for k and n-k.

The first major result consists, as the above one, from two formula iterations:

Asymptotic expansion

a)

[;\frac{F_{2 \cdot m}}{F_{2 \cdot m + 1}} \cdot n <k< \frac{F_{2 \cdot m - 1}}{F_{2 \cdot m}} \cdot n \Rightarrow;]

[;\tau_1(n,k) = (\sum_{i=1}^{2 \cdot m - 1}(-1)^{i+1} \cdot F_{i}) \cdot n + (\sum_{j=2}^{2\cdot m} (-1)^{j+1} \cdot F_{j}) \cdot k +;]

[;\tau_1(F_{2 \cdot m-1} \cdot k - F_{2\cdot m - 2} \cdot n, F_{2 \cdot m - 1} \cdot n - F_{2 \cdot m} \cdot k);]

b)

[;\frac{F_{2 \cdot m}}{F_{2 \cdot m + 1}} \cdot n <k< \frac{F_{2 \cdot m + 1}}{F_{2 \cdot m + 2}} \cdot n \Rightarrow;]

[;\tau_1(n,k) = (\sum_{i=1}^{2 \cdot m }(-1)^{i+1} \cdot F_{i}) \cdot n + (\sum_{j=2}^{2\cdot m + 1} (-1)^{j+1} \cdot F_{j}) \cdot k +;]

[;\tau_1(F_{2 \cdot m - 1} \cdot n - F_{2 \cdot m} \cdot k, F_{2 \cdot m+1} \cdot k - F_{2\cdot m } \cdot n);]

where 

[;F_{1} = F_{2}=1;]

[;F_{m+2}=F_{m+1} + F_{m};] 

stand for the terms of the Fibonacci sequence, and by natural convention:

[;F_{0} = 0;]

The proof follows by mathematical induction, the base case being ensured by the outer row recurrence  (for a)) and the intermediate recurrence (for b)). For the inductive step we will suppose that a) and b) hold for n arbitrarily fixed m. Through the outer row recurrence results that a) holds for m+1 as well and in a similar fashion b) holds for m+1 as well, thus concluding the inductive reasoning.
Since from the previous part we have that:

[;\frac{\tau_2(n)}{n} < 1;]

and for a conveniently chosen m, via the asymptotic expansion we have that:

[;\frac{\tau_2(n)}{n} > \frac{F_{2\cdot m + 1} - 1}{ F_{2 \cdot m + 1}};]

we can now state that:

[;(\frac{\tau_2(n)}{n})_{n \ge 1} \rightarrow 1;]

There is also another gain (from the asymptotic expansion), from the perspective of the positional peak: if we squeeze k such that the fraction n/k becomes closer and closer to the golden ratio we get values of the euclidean summation pretty close to n. Alas we can only conjecture for now that:

[;(\frac{\tau_3(n)}{n})_{n \ge 1} \rightarrow \phi;]

where:

[;\phi = \frac{1}{\varphi} \approx 0.618;]

is called the golden ratio conjugate.

Next: Fractal candidate building blocks

Discrete connections (part III)

For the positional peak, its boundaries are straightforward (definition, inner row  recurrence and lower boundary):

[; \frac{n}{2}<\tau_3(n)<n \; \forall \; n \ge n_{\tau_3};]

For the peak, it's a two step argument. For the lower bound we consider two cases: n odd and respectively n even. For every such case we take a conveniently chosen pair, compute the euclidean summation and truncate it up to a suitable value:

 [; n = 2 \cdot k + 1 \Rightarrow ;]

[; \tau_1(n,k+2) = n-(k+2) + \tau_1(k+2, k-1) \Rightarrow ;]

[; \tau_1(n,k+2) \ge k-1 + 3= k+2 > k + \frac{1}{2} = \frac{n}{2} ;]

[; n=2\cdot k \Rightarrow ;]

[;\tau_1(n,k+1) = n-(k+1) + \tau_1(k+1, k-1) \Rightarrow ;]

[; \tau_1(n,k+1) \ge k-1 + 2= k+1 > k = \frac{n}{2};]

Here also the inequality holds from a rank on, due to the approximations used.

For the upper bound, our reasoning will follow the rules of the mathematical induction:

 base case

[;\tau_1(1,0)=0<1;]

inductive step (n arbitrarily fixed)

[;\tau_1(k,r)<r \; \forall \; k,r \; such \; that \; n>k>r \Rightarrow ;]

[;\forall \; k<n \Rightarrow \exists \; q,r \; such \; that \;q \ge 1, k>r, n=q\cdot k + r  \Rightarrow ;]

[;\tau_1(n,k) = r + \tau_1(k,r) < r + k \le r + q \cdot k = n;]

thus concluding our reasoning.

We get in conclusion:

[; \frac{n}{2}<\tau_2(n)<n \; \forall \; n \ge n_{\tau_2};]

Next: Asymptotic behaviors of peak and positional peak.