# Hyperanalytic function

## Introduction

An hyperanalytic function is a function that is locally given by a converging power series with a speed that corresponds to tetration.

## Definitions

Let $V$ and $W$ be complete Hausdorff topological vector spaces, let $W$ be locally convex, let $c$ be an element of $V$ , and let (a_0,a_1,a_2,$\ldots$ ) be an infinite sequence of homogeneous operators from $V$ to $W$ with each $a_k$ of degree $k$ .

Given an element $c$ of $V$ , consider the infinite series $\sum_k a_k(x - c)^k$ (a power series). Let $U$ be the interior of the set of $x$ such that this series converges in $W$ ; we call $U$ the domain of convergence of the power series. This series defines a function from $U$ to $W$ ; we are really interested in the case where $U$ is inhabited, in which case it is a balanced neighbourhood of $c$ in $V$ (which is Proposition 5.3 of [Bochnak--Siciak](#BS)).

Let $D$ be any subset of $V$ and $f$ any continuous function from $D$ to $W$ . This function $f$ is hyperanalytic if, for every $c \in D$ , there is a power series as above with inhabited domain of convergence $U$ such that $f(x) = \sum_k a_k(x - c)^k$ for every $x$ in both $D$ and $U$ and just rare $a_k$ are not equal zero. (That $f$ is continuous follows automatically in many cases, including of course the finite-dimensional case.)

### Examples

It is known that there is a fundamental connection between analyticity of the function and the convergence of its Fourier coefficients. The better the function, the faster its coefficients tend to zero, and vice versa. The power decrease of Fourier coefficients is inherent in functions of the $C^{k}$ class while exponential to analytical functions. Here there is a possibility of existence of the hyperanalytic functions, for which the decrease of the Fourier coefficients corresponds to tetration.

Natural hyperanalytic function occurs when considering reticulum with a step $L$ , in which nodes there are not defined yet objects. The distribution of center's objects can be described using the reticulum functions (RF). The definition of a one-dimensional RF is based on the following identity: $$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x}{\sigma})^{2}}dx=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\frac{L}{2}}^{\frac{L}{2}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-nL}{\sigma})^{2}}dx=1. \label{e1}$$$

#### Definition

From here RF is $\mathbb{R}(x)=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-nL}{\sigma})^{2}}$

It is obvious that the RF can not be laid out in the Fourier series because it does not have antiderivative that can be expressed as elementary functions. By virtue of this RF cannot be decomposed into even and odd functions, while an arbitrary analytic function $f$ can be only presented in the form of sum of odd and even functions in the interval $[a,b]$ : $f\left(x\right)=g\left(x\right)+h\left(x\right),$ where $g\left(x\right)=\frac{f\left(x-a\right)-f\left(b-x\right)}{2},$ $h\left(x\right)=\frac{f\left(x-a\right)+f\left(b-x\right)}{2}.$

Due to this the RF can be laid out in an endless row of two primitive hyperanalytic functions by sequential attempts to decompose on even and odd functions. Thus, the RF can be decomposed by the simplest way, but such a series is not one like the orthonormal basis of Fourier series.

### Decomposition of RF

#### Definitions

$\mathbb{R}\left(0\right)$ is $\mathbb{R}\left(0\right)=\mathbb{R}_{max}=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}\left(\frac{-n}{\sigma}\right)^{2}}.$

$\mathbb{R}\left(1/2\right)$ is $\mathbb{R}\left(1/2\right)=\mathbb{R}_{min}=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}\left(\frac{1/2-n}{\sigma}\right)^{2}}.$

Then $A_{0}$ is the mean value of RF: $A_{0}=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}.$

As it follows from (1) the mean value of RT is 1. However as will be seen from the further, it is expedient to choose the greater value of the decomposition's constant member. Introduce the following definitions:

### First difference

One can approximate first difference by the following way: $A_{1}\left(x\right)=\frac{\mathbb{R}_{max}-\mathbb{R}_{min}}{2}\cos\left(2\pi x\right).$

#### Definition

Let introduce parameter of the fine structure $\alpha$ as function of $\sigma$ : $\alpha\left(\sigma\right)=\frac{1}{2}\frac{\mathbb{R}_{max}-\mathbb{R}_{min}}{\mathbb{R}_{max}+\mathbb{R}_{min}}.$

Now $A_{1}\left(x\right)$ can be expressed as: $A_{1}\left(x\right)=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}\left(2\alpha\left(\sigma\right)cos\left(2\pi x\right)\right).$ The choice of the name and symbol of this parameter is due to the fact that $\alpha\left(0.4992619105929628\right)=\alpha=\frac{e^{2}}{4\pi\epsilon_{0}\hbar c}$

is the value known in physics as a fine structure constant.

#### Even differences

Even differences are a primitive hyperanalytic function $\overline{\mathbb{V}}(2i\times2\pi x)$ , which is quasisymmetric relative to the point $\text{x=0.25}$ .

Its symmetrical part approximated in the following way: $A_{2i}\left(x\right)=c_{2i}\left(cos\left(2i\times2\pi x\right)-1\right)$

and $\sum_{i=1}^{\infty}c_{2i}=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}-1=2 * \sum_{k=1}^{\infty} \alpha^{4^{k}}$

Using the value $\mathbb{R}\left(1/4\right)=\mathbb{R}_{1/4}=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}\left(\frac{1/4-n}{\sigma}\right)^{2}}$

define the amplitude for $c_{2}$ : $\frac{1}{2}\left(\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}-\mathbb{R}_{1/4}\right)=2\alpha^{4}.$

This definition allows to select approximation $A\left(x\right)$ in the form: $A\left(x\right)=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}\left(1+2\alpha\left(\sigma\right)cos\left(2\pi x\right)\right)+2\alpha^{4}\left(cos\left(2\times2\pi x\right)-1\right).$

#### Odd differences

Odd differences are a primitive hyperanalytic function $\mathbb{W}\left((2i-1)\times2\pi x\right)$ which is quasiantisymmetric relative to the point $\text{x=0.25}$ .

Quasiantisymmetry of $\mathbb{W}\left(2\pi x\right)$ follows from the fact that the integral of $A\left(x\right)$ differs from 1: $\int_{-1/2}^{1/2}\text{A}\left(x\right)\text{dx}-1=\frac{1}{4}\left(\mathbb{R}_{max}+\mathbb{R}_{min}\right)+\frac{1}{2}\mathbb{R}_{1/4}-1\simeq1.02E-34.$

Thus function $\mathbb{W}\left((2i-1)\times2\pi x\right)$ should be decomposed in the even and odd function. Its even part is: $\mathbb{W}^{\text{qs}}\left((2i-1)\times2\pi x\right)=\frac{\mathbb{W}\left((2i-1)\times2\pi x\right)+\mathbb{W}\left((2i-1)\times2\pi\left(0.5-x\right)\right)}{2}=\overline{\mathbb{V}}(2(i+1)\times2\pi x).$

However, as shown above, $overline{\mathbb{V}}(2i\times2\pi x$ ) is not an even function.

The odd part of $\mathbb{W}\left((2i-1)\times2\pi x\right)$ is no longer a hyperanalytic function and is equal to: $W^{\text{qa}}\left((2i-1)\times2\pi x\right)=\frac{\mathbb{W}\left((2i-1)\times2\pi x\right)-\mathbb{W}\left((2i-1)\times2\pi\left(0.5-x\right)\right)}{2}.$

It can be approximated with any degree of accuracy following way: $A(W^{\text{qa}}\left((2i-1)\times2\pi x\right))=\beta(cos\left(3(2i-1)\times2\pi x\right)-cos\left((2i-1)\times2\pi x\right)),$

where $\beta$ is a normalizing multiplier.

Thus, the approximation of $\mathbb{R}(x)$ is: $A\left(x\right)=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}(1+2\alpha cos\left(2\pi x\right)) +2\sum_{i=1}^{\infty}\alpha^{4^{i}}\left(cos\left(2i\times 2\pi x\right)-1\right)+\frac{2}{\mathbb{W}_{max}}\sum_{i=1}^{\infty}\alpha^{9{i}^2}\left(cos\left(3 \times (2i-1)\times 2\pi x\right)-cos\left((2i-1) \times 2\pi x\right)\right),$

where $\mathbb{W}_{max}$ is a normalizing multiplier.

### Three-dimensional RF

Three-dimensional RF $\mathbb{R}\left(x,y,z\right)$ can be obtained from the definition $\left(1.2\right)$ : $\mathbb{R}\left(x,y,z\right)=\mathbb{R}_{max}^{2}\mathbb{R}\left(x\right).$

Thus, the approximation of the three-dimensional RF is also the series of the fine structure constant $\alpha$ along any axis of the reticulum three-dimensional space, and the constant itself is a function of the dimensionless parameter $\sigma$ , which is equal to quotient of the "diameter" of some physical object, located in each cell, to the grid step $L$ [/itex].

### Quantum derivative with respect to time

To quantize the time the direct use of the lattice idea is too formal. It is therefore appropriate to use a definition of derivative with respect to time but without moving to the limit. Let $\mathbb{R}\left(t\right)$ is RF on a unit interval $\left[-\text{T/2},\text{T/2}\right]$ and $\tau=\sigma$ и $\text{T}=1$ : $\mathbb{R}\left(t\right)=\frac{1}{\tau\sqrt{2\pi}}\sum_{i=-\infty}^{\infty}\left[\exp\left(-\frac{1}{2}\left(\frac{t+\text{T/4}-i}{\tau}\right)^{2}\right)-\exp\left(-\frac{1}{2}\left(\frac{t-\text{T/4}-i}{\tau}\right)^{2}\right)\right].$

By consistently subtracting sinuses, one can show that the approximation of the $\mathbb{R}\left(t\right)$ has the following form: $A\left(t\right)=\sum_{k=0}^{\infty}\left(-1\right)^{k+1}a_{k}sin\left(2\pi\left(2k+1\right)t\right).$

Let use $k+1$ equations with different values of $l$ to determine the coefficient's values $a_{k}$ : $\sum_{i=0}^{k}\left(-1\right)^{i}a_{i}sin\left(\frac{2i+1}{2l+1}\frac{2\pi}{4}\right)=\mathbb{R}\left(\frac{1}{4\left(2l+1\right)}\right).$

Given that A\left(1/4\right) is numerically equal to $2\left(\mathbb{R}_{max}\left(\tau\right)+\mathbb{R}_{min}\left(\tau\right)\right)\alpha\left(\tau\right)$ , equation can be written as follows: $\alpha_{eff}\left(t,\tau\right)=\frac{1}{2\left(\mathbb{R}_{max}\left(\tau\right)+\mathbb{R}_{min}\left(\tau\right)\right)}\sum_{k=0}^{\infty}\left(-1\right)^{k+1}a_{k}sin\left(2\pi\left(2k+1\right)t\right).$

$\mathbb{R}\left(t\right)$ is also a hyperanalytic function, as the next approximation takes place: $\alpha_{eff}\left(t,\tau\right)=\sum_{k=0}^{\infty}\left(-1\right)^{k+1}\alpha^{(2k+1)^{2}}sin\left(2\pi\left(2k+1\right)t\right).$

The theory of hyperanalytic function was constructed to some extent by A. Rybnikov (2014) http://www.gaussianfunction.com/.