Discrete connections (part XI)

This part can be better classified as a "bigger picture" overview. We've seen in the past articles several strong hints concerning the global properties of the euclidean summation but no word yet about local ones. Here we try to give some of these local properties a go, while raising subtle bridges between the euclidean summation and the euclidean fractal.

As accustomed, we will define some derivatives of the euclidean summation peak and euclidean summation positional peak:

[; \tau_2^{p}(n) = \left\{ \begin{array}{l} 0, \; if \; I_n^{p} = \emptyset \\ \max\{ \tau_1(n,k) \; | \; k \in I_{n}^{p} \}, \; otherwise \end{array} \right. ;]

[; \tau_3^{p}(n) = \left\{ \begin{array}{l} 0, \; if \; I_{n}^{p} = \emptyset \\ \max\{ k \in I_{n}^{p} \; | \; \tau_1(n,k) = \tau_2^{p}(n) \}, \; otherwise \end{array} \right. ;]

where:

[; p \in \mathbf{N}, \; p \ge 1 ;]

[; I_{n}^{p} = \{k \in \mathbf{N} \; | \; \frac{n}{p+1} \le k \le \frac{n}{p} \} ;]

The target conjectures for these derivatives are:

[; \lim_{n \rightarrow \infty} \frac{\tau_2^{p}(n)}{n} = \frac{\varphi}{p + \phi} ;]

[; \lim_{n \rightarrow \infty} \frac{\tau_3^{p}(n)}{n} = \frac{1}{p+\phi} ;]

For p=1 these conjectures translate to the result proven for the euclidean summation peak and respectively to the conjecture for the euclidean summation positional peak.

We can go even deeper with this construction, and involve yet another classic: continued fractions (and thus tighten the links with the euclidean fractal). Indeed:

[; \forall s \in \mathbf{N}, \; s > 0 \; and \; \forall w = (p_1, \ldots, p_s) \in (\mathbf{N} \setminus \{0\})^{s} ;]

let us define:

[; \tau_2^{w}(n) = \left\{ \begin{array}{l} 0, \; if \; I_n^{w} = \emptyset \\ \max\{ \tau_1(n,k) \; | \; k \in I_{n}^{w} \}, \; otherwise \end{array} \right. ;]
[; \tau_3^{w}(n) = \left\{ \begin{array}{l} 0, \; if \; I_{n}^{w} = \emptyset \\ \max\{ k \in I_{n}^{w} \; | \; \tau_1(n,k) = \tau_2^{w}(n) \}, \; otherwise \end{array} \right. ;]

where

[; I_{n}^{w} = \{k \in \mathbf{N} \; | \; \cfrac{n}{p_1+\cfrac{1}{\ddots + \cfrac{1}{p_s + \epsilon(s)}}}} \le k \le \cfrac{n}{p_1+\cfrac{1}{\ddots + \cfrac{1}{p_s + 1 - \epsilon(s)}}}} \} ;]

and:

[;\epsilon(s) = \left\{ \begin{array}{l} 0, \; if \; s \; even \\ 1, \; otherwise \end{array} \right. ;]

The conjectured results for these generalized derivatives are:

[; \lim_{n \rightarrow \infty} \frac{\tau_3^{w}(n)}{n} = \cfrac{1}{p_1+\cfrac{1}{\ddots + \cfrac{1}{p_s + \phi }}} ;]

Next: To be continued ...

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 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 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