Discrete connections (part X)

From this part on we strive to define candidates for the left and right mappings from both the euclidean summation and the euclidean fractal point of view.
Let us define by the left and right mappings of the normalized euclidean summation the functions conjectured to have the following properties:

D)

[;\varepsilon^{\tau}_{l}, \varepsilon^{\tau}_{r}:(0,1) \rightarrow [0,1];]
L)
[;\varepsilon^{\tau}_{l}(x)=\lim_{n \rightarrow \infty}\frac{\tau_1(n,\lceil x \cdot n \rfloor)}{n};]
R)
[;\varepsilon^{\tau}_{r}(x)=\lim_{n \rightarrow \infty}\frac{\tau_1(n,\lfloor x \cdot n \rceil)}{n};]

where:

[;\lceil x \rfloor = \lceil x \rceil - 1;]

[;\lfloor x \rceil = \lfloor x \rfloor + 1;]

Given the properties of the euclidean summation, for the moment we can prove only:

- that D) is well defined, since:

[;0 < \tau_1(n,k) < n;]

and both the lower and the upper bounds are reached;

- that the sequences in L) and R), due to their boundedness, both contain convergent subsequences.

We can of course prove the above for discrete cases:

[; x = \frac{1}{p}, \; where \; p \in \mathbf{N}, p \ge 2 ;]

For proving L, if p divides n we have:

[; n = k \cdot p \Rightarrow \lceil n \cdot x \rfloor = k - 1 \Rightarrow n = p \cdot (k-1) + p ;]

which becomes a valid euclidean division (with quotient p and remainder p), if  k-1>p. Since we are dealing with a limit, this assumption is safe.

Now on the other hand we get:

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

[; 0 < \tau_1(n, k-1) < p + \max\{r + \tau_1(p,r) | r \in \mathbf{N}, r < p \} \Rightarrow;]

[; 0< \tau_1(n, k-1) < 3 \cdot p \Rightarrow \lim_{n \rightarrow \infty, \; p|n}} \frac{\tau_1(n, \lceil n \cdot \frac{1}{p} \rfloor)}{n} = 0 ;]

If p does not divide n then we have:

[; n = k \cdot p + r, \; where \; r \in \mathbf{N}, \; 0<r<p \Rightarrow ;]

[; \lceil n \cdot x \rfloor = k \Rightarrow n = p \cdot k + r ;]

which becomes a valid euclidean divison if k>p.

In a similar fashion as above we get:

[; 0 < \tau_1(n,k) < 3 \cdot r \Rightarrow 0 < \tau_1(n,k) < 3 \cdot p \Rightarrow ;]

[; \lim_{n \rightarrow \infty, p \nmid n} \frac{\tau_1(n, \lceil n \cdot \frac{1}{p} \rfloor)}{n} = 0 ;]

For proving R we can use the general case:

[; n = k \cdot p + r, \; where \; r \in \mathbf{N}, \; 0 \le r < p \Rightarrow ;]

[; \lfloor n \cdot x \rceil = k+1 \Rightarrow n = (p-1) \cdot (k+1) + k + 1 - p + r ;]

which becomes a valid euclidean division if k>p.

Therefore:

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

If we raise the boundary of k we get:

[; k > 2 \cdot p \Rightarrow ;]

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

[; k + 1 \le \tau_1(n, k + 1) < k + 1 + 2 \cdot (p - r) \Rightarrow ;]

[; k + 1 \le \tau_1(n, k + 1) < k + 1 + 2 \cdot p \Rightarrow ;]

[; \lim_{n \rightarrow \infty} \frac{\tau_1(n, \lfloor n \cdot \frac{1}{p} \rceil )}{n} = \frac{1}{p} ;]

Thus we have proven not only that L and R are well defined for x=1/p but that the limits obtained make some of the conjectured properties of the left and right mappings true.

Next: One step deeper ...

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.