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 V)

This part will deal mainly with the constructive elements of a fractal class derived from the conjectured asymptotic nature of the euclidean summation.
First some basic sequences and functions:

r-golden ratio builder

[;g_0^r=r, \; g_{n+1}^r = 1 + \frac{1}{g_n^r};]

upward shifted by r identity

[;u_n^r=n+r;]

upper ratio coefficient

[;\overline{\lambda}:[0,\infty) \rightarrow (0,1], \; \overline{\lambda}(x)=\frac{1}{1+x};]

where r>0 is a real number.

Let us consider some properties of the above sequences and functions:

[;\lim_{n\rightarrow \infty}\overline{\lambda}(g_n^r) = 1-\phi, \sum_n (-1)^n \cdot \prod_{i=0}^{n} \overline{\lambda}(g_i^r) \; convergent;]

[;\lim_{n \rightarrow \infty} \overline{\lambda}(u_n^r) = 0, \prod_n (1 - \overline{\lambda}(u_n^r)) \; convergent \; to \; 0;]

For the upward shifted by r identity, the conclusions are self evident due to:

[;\overline{\lambda}(u_n^r) = \frac{1}{1+n+r};]

and

[;\prod_{i=0}^{n} (1 - \overline{\lambda}(u_n^r)) = \frac{r}{1+n+r};]

For the r-golden ratio builder, the first conclusion's proof is based upon the relative positions of r and the golden ratio.

Indeed, if r is equal to the golden ratio then the r-golden ratio builder is stationary (and equal to the golden ratio).

From the two left cases let's consider the case r is smaller than the golden ratio. With this assumption we get:

[;r^2 -r -1 < 0 \Rightarrow g_1^r - g_0^r = -\frac{r^2-r-1}{r} > 0;] 

By means of mathematical induction one can prove that the subsequences of even index and odd index terms of the r-golden ratio are respectively increasing and decreasing. Since both sequences fall between the two first terms of the sequence results that they are both convergent to respectively g and h. 

We get successively:

[;\left\{ \begin{array}{rcl} g & = &1+\frac{1}{h}\\ h & = &1+\frac{1}{g} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{rcl} g & = &1+\frac{1}{1+\frac{1}{g}} \\ h & = & 1+\frac{1}{g} \end{array} \right. \Leftrightarrow;]
[;\left\{ \begin{array}{rcl}g & = & \frac{2\cdot g +1}{g+1} \\ h & = & 1 + \frac{1}{g}\end{array}\right. \Leftrightarrow \left\{ \begin{array}{rcl} g^2 -g -1 & = & 0\\ h & = & 1 + \frac{1}{g} \end{array} \right. \Leftrightarrow;]
[;\left\{ \begin{array}{rcl}g & = & \varphi \\ h & = & 1 + \frac{1}{g}\end{array}\right. \Leftrightarrow \left\{ \begin{array}{rcl} g & = & \varphi \\ h & = & \varphi \end{array} \right.;]

Therefore the entire sequence is convergent to the golden ratio. Since the golden ratio is not equal to 1 results that:

[;\sum_n (-1)^n \cdot \prod_{i=0}^{n} \overline{\lambda}(g_i^r);]

is convergent.

Having these sequences and functions completely described let us consider the two mechanisms we will use in order to define the previously mentioned class of fractals:

The top value base sequence

For any four complex numbers u,v,w,z shaping a trapezoid and the sequences:

[;(\alpha_n)_{n\ge 0}, \; (\beta_n)_{n\ge 0};]

defined by:

[;\alpha_0=u, \; \beta_0=v, \beta_1=w,\; \alpha_1=z;]

and

[;\left\{ \begin{array}{rcl} \alpha_{n+2} & = & (1 - \overline{\lambda}(g_n^r))\cdot \alpha_{n+1}+ \overline{\lambda}(g_n^r) \cdot \alpha_{n} \\ \beta_{n+2} & = & (1 - \overline{\lambda}(g_n^r))\cdot \alpha_{n+1}+ \overline{\lambda}(g_n^r) \cdot \beta_{n} \end{array} \right. ;]

we have that:

[;(\alpha_n)_{n\ge 0} \rightarrow \alpha;]

The self-similarity base sequence

For any four complex numbers u,v,w,z shaping a trapezoid and the sequences:

[;(\tilde{\alpha}_n)_{n\ge 0}, \; (\tilde{\beta}_n)_{n\ge 0};]

defined by:

[;\tilde{\alpha}_0=u, \; \tilde{\beta}_0=v, \tilde{\beta}_1=w,\; \tilde{\alpha}_1=z;]

and

[;\left\{ \begin{array}{rcl} \tilde{\alpha}_{n+2} & = & (1 - \overline{\lambda}(u_n^r))\cdot \tilde{\alpha}_{n+1}+ \overline{\lambda}(u_n^r) \cdot \tilde{\alpha}_{0} \\ \tilde{\beta}_{n+2} & = & (1 - \overline{\lambda}(u_n^r))\cdot \tilde{\alpha}_{n+1}+ \overline{\lambda}(u_n^r) \cdot \tilde{\beta}_{0} \end{array} \right. ;]

we have that:

[;(\tilde{\alpha}_n)_{n\ge 0} \rightarrow \tilde{\alpha}_0;]

For both sequences r is the scaling ratio of the two parallel edges of the trapezoid i.e.:

[;(u-v)=r\cdot (z-w);]

The construction of the above two sets of sequences is better depicted by the following figures:

and

Next: The euclidean fractal class

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

Golden ratio and life (part I)

Warning: Starting with this blog you may need some additional magic: in order to typeset mathematical formulas I used a script file named TeX THE WORLD. Consequently you may have to enable javascript in your browser (Greasemonkey and the same user script). Sorry about the inconvenience, but this way maths look better.

As it's name states, the golden ratio has everything to do with geometry and aesthetics. It was defined by the following property: "Given a line segment of length c, how can we divide it in two sub-segments of lengths a and b (c=a+b, a > b) such that the ratio of the large over the small equals the ratio of the segment over the large?" Otherwise said:

[; \varphi = \frac{a}{b} = \frac{c}{a} ;]

Now, replacing c by the sum of a and b, we get:

[; \varphi = \frac{1 + \varphi}{\varphi} ;]

hence:

 [; \varphi = \frac{1+\sqrt{5}}{2} \approx 1.618 ;]

This proportion was deeply engraved in ancient (the Parthenon), Renaissance (Mona Lisa) and modern (The Sacrament of the Last Supper) masterpieces of painters and architects, and all for a reason: mathematics ARE aesthetically pleasing.
And coincidences don't stop here: remember the Fibonacci sequence? Let's consider consecutive terms ratios and place them like this:

[;\frac{1}{1}, \frac{3}{2}, \frac{5}{8}, ... ? ... , \frac{13}{8},\frac{5}{3},\frac{2}{1};]

We get two sequences, an increasing one and a decreasing one, both leading to ... the golden ratio. So, if you want to use golden ratio-like proportions, the Fibonacci numbers are your best choice, no need for tedious square roots.