# Tetration Forum

Full Version: New mathematical object - hyperanalytic function
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New mathematical object - hyperanalytic function is introduced. The convergence of hyperanalytic functions is substantially above than the convergence of analytic functions. A specific sample of hyperanalytic function is the reticulum function (RF). This function describes the reticulum space-time. RF can't be decomposed into the Fourier series and, therefore, RF does not provide the conservation of parity as the analytic functions do. Thanks to this, the RF can be decomposed in an endless series of two primitive hyperanalytic functions by sequential attempts of decomposition in the even and odd functions. The unique parameter of such series is the fine structure constant.
It allows combine all fundamental interactions into the Naturally-Unified Quantum Theory of Interactions. The price of such quantum unification is the reticulum space-time.

How is this related to tetration?
(12/31/2019, 11:04 PM)bo198214 Wrote: [ -> ]How is this related to tetration?

Directly.
For example, function

Code:
\mathbb{R}(x)=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-nL}{\sigma})^{2}}

has approximation

Code:
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

Code:
\alpha\left(\sigma\right)=\frac{1}{2}\frac{\mathbb{R}_{max}-\mathbb{R}_{min}}{\mathbb{R}_{max}+\mathbb{R}_{min}}.
(01/01/2020, 10:51 PM)arybnikov Wrote: [ -> ]
(12/31/2019, 11:04 PM)bo198214 Wrote: [ -> ]How is this related to tetration?

Directly.
For example, function
$\mathbb{R}(x)=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-nL}{\sigma})^{2}}$

has approximation

$A\left( x \right)=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}(1+2\alpha cos(2\pi x )$

$+2\sum_{i=1}^{\infty} \alpha^{4^i}( \cos ( 2i \times 2\pi x ) -1 )
+\frac{2}{\mathbb{W}_{max}}\sum_{i=1}^{\infty}\alpha^{9{i}^2}\left( cos\left(3 \times (2i-1)\times 2\pi x \right)
-cos\left( (2i-1) \times 2\pi x \right) \right)$
,

where

$\alpha\left(\sigma\right)=\frac{1}{2}\frac{\mathbb{R}_{max}-\mathbb{R}_{min}}{\mathbb{R}_{max}+\mathbb{R}_{min}}$.
(01/02/2020, 12:15 AM)Daniel Wrote: [ -> ]
(01/01/2020, 10:51 PM)arybnikov Wrote: [ -> ]
(12/31/2019, 11:04 PM)bo198214 Wrote: [ -> ]How is this related to tetration?

Directly.
For example, function
$\mathbb{R}(x)=\frac{1}{\sigma\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-nL}{\sigma})^{2}}$

has approximation

$A\left( x \right)=\frac{\mathbb{R}_{max}+\mathbb{R}_{min}}{2}(1+2\alpha cos(2\pi x ))$

$+2\sum_{i=1}^{\infty} \alpha^{4^i}( cos(2i \times 2 \pi x )-1)+\frac{2}{\mathbb{W}_{max}}\sum_{i=1}^{\infty}\alpha^{9i^2}\left( cos\left( 3 \times (2i-1) 2\pi x \right) - cos\left( (2i-1) 2\pi x \right) \right)$,

where

$\alpha\left(\sigma\right)=\frac{1}{2}\frac{\mathbb{R}_{max}-\mathbb{R}_{min}}{\mathbb{R}_{max}+\mathbb{R}_{min}}$.

Unfortunately I still have problem with $\rightarrow$ and [?].