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 ...