<![CDATA[Tetration Forum - Mathematical and General Discussion]]> https://tetrationforum.org/ Tue, 12 May 2026 11:43:12 +0000 MyBB <![CDATA[Recurrence relations and differential equations]]> https://tetrationforum.org/showthread.php?tid=1819 Wed, 06 May 2026 06:58:49 +0000 Natsugou]]> https://tetrationforum.org/showthread.php?tid=1819 I'm trying to generalize recurrence relations and autonomous differential equations.
I think that \(\mathbb{R}_+\)-actions are the straightforward continuous version of \(\mathbb{N}\)-actions, so I checked whether an orbit of a \(\mathbb{R}_+\)-action is always a solution (an integral curve) of an autonomous differential equation, and the opposite.

\(\gamma: \mathbb{R} \to \mathbb{R}^n\) is an orbit of a \(\mathbb{R}\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u \in \mathbb{R}\).
\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u > 0\).
\(\gamma\) is a solution of an autonomous differential equation \(\gamma'(t) = F(\gamma(t))\) if and only if \(\gamma(s) = \gamma(t) \implies \gamma'(s) = \gamma'(t)\). (Define \(F(x) := \gamma'(t)\) when \(x = \gamma(t)\). The \(F\) is well-defined on \(\gamma(\mathbb{R})\) because of the hypothesis.)

From the Picard–Lindelöf theorem, if \(F: \mathbb{R}^n \to \mathbb{R}^n\) is Lipschitz continuous, a solution \(\gamma: \mathbb{R} \to \mathbb{R}^n\) of a differential equation \(\gamma'(t) = F(\gamma(t))\) is always an orbit of a \(\mathbb{R}\)-action.
Trivially an orbit of a \(\mathbb{R}\)-action is an orbit of a \(\mathbb{R}_+\)-action.
If an orbit of a \(\mathbb{R}_+\)-action is differentiable,  it's always a solution of an autonomous differential equation. (The proof is described below.)
Therefore let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, then
\[
  \begin{array}{ll}
    & \text{there exists a Lipschitz continuous \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)} \\
    \implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}\)-action} \\
    \implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action} \\
    \implies & \text{there exists \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)}.
  \end{array}
\]
The opposites are not true.

An orbit of a \(\mathbb{R}\)-action that is not a solution of an autonomous differential equation of a Lipschitz continuous function:
Since
\[
  \gamma(t) =
  \begin{cases}
    -t^2 & (t \leq 0) \\
    t^2 & (t \geq 0)
  \end{cases}
\]
is injective, the \(\gamma: \mathbb{R} \to \mathbb{R}^1\) is an orbit of \(\mathbb{R}\)-action.
Assume there exists a Lipschitz continuous \(F: \mathbb{R}^1 \to \mathbb{R}^1\) such that \(\gamma'(t) = F(\gamma(t))\).
Then there exists \(L \in \mathbb{R}\) such that \(|\gamma'(t_1) - \gamma'(t_2)| = |F(\gamma(t_1)) - F(\gamma(t_2))| \leq L|\gamma(t_1) - \gamma(t_2)|\) for all \(t_1, t_2 \in \mathbb{R}\).
Let \(t_1 \neq 0\) and \(t_2 = 0\).
We see \(\gamma'(t) = 2|t|\), so \(2|t_1| \leq L|t_1^2|\). Therefore
\[
  \left|\frac{1}{t_1}\right| = \left|\frac{t_1}{t_1^2}\right| \leq \frac{L}{2}.
\]
\(\left|\frac{1}{t_1}\right| \longrightarrow \infty\) when \(t_1 \longrightarrow 0\), that is a contradiction.

A differentiable orbit of a \(\mathbb{R}_+\)-action that is not an orbit of a \(\mathbb{R}\)-action:
\[
  \gamma: \mathbb{R} \to \mathbb{R}^1, \quad
  \gamma(t) =
  \begin{cases}
    t^2 & (t \leq 0) \\
    0 & (t \geq 0)
  \end{cases}
\]
is an example of it.

A solution of an autonomous differential equation that is not an orbit of a \(\mathbb{R}_+\)-action:
\[
  \gamma: \mathbb{R} \to \mathbb{R}^1, \quad
  \gamma(t) =
  \begin{cases}
    0 & (t \leq 0) \\
    t^2 & (t \geq 0)
  \end{cases}
\]
is a solution of \(\gamma'(t) = F(\gamma(t))\) where
\[
  F: \mathbb{R}^1 \to \mathbb{R}^1, \quad
  F(x) =
  \begin{cases}
    0 & (x \leq 0) \\
    2\sqrt{x} & (x \geq 0),
  \end{cases}
\]
while \(\gamma(-1) = \gamma(0)\) and \(\gamma(-1+1) = 0 \neq 1 = \gamma(0+1)\).

Proof of that a differentiable orbit of a \(\mathbb{R}_+\)-action is a solution of an autonomous differential equation:
We'll prove the contraposition.
Let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, \(\gamma(t) = (\gamma_1(t), \dots, \gamma_n(t))\). (\(\gamma'(t)\) denotes \((\gamma_1'(t), \dots, \gamma_n'(t))\).)
Assume \(\gamma\) is not a a solution of an autonomous differential equation.
Then there are \(t_1, t_2 \in \mathbb{R}\) such that \(\gamma(t_1) = \gamma(t_2)\) and \(\gamma'(t_1) \neq \gamma'(t_2)\).
Therefore \(\gamma'_i(t_1) \neq \gamma'_i(t_2)\) for some \(i\).
Define \(x = \gamma_i(t_1) = \gamma_i(t_2)\), \(\alpha = \gamma_i'(t_1)\), and \(\beta = \gamma_i'(t_2)\). We can assume \(\alpha < \beta\) w.l.o.g.
We know
\[
  \forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall h \in (-\delta_1, \delta_1), \left|\frac{\gamma_i(t_1+h)-x}{h} - \alpha\right| < \varepsilon_1
\]
and
\[
  \forall \varepsilon_2 > 0, \exists \delta_2 > 0, \forall h \in (-\delta_2, \delta_2), \left|\frac{\gamma_i(t_2+h)-x}{h} - \beta\right| < \varepsilon_2.
\]
Thus
\[
  x + \alpha h - \varepsilon_1|h| < \gamma_i(t_1+h) < x + \alpha h + \varepsilon_1|h|,
\]
\[
  x + \beta h - \varepsilon_2|h| < \gamma_i(t_2+h) < x + \beta h + \varepsilon_2|h|.
\]
Since \(\beta - \alpha > 0\) there are \(\varepsilon, \varepsilon' \in \mathbb{R}\) such that \(0 < \varepsilon < \varepsilon' < \beta - \alpha\).
Substitute \(\varepsilon_1 = \varepsilon\) and \(\varepsilon_2 = \beta - \alpha - \varepsilon'\), then for any \(0 < h < \min\{\delta_1, \delta_2\}\),
\begin{align*}
  \gamma_i(t_1+h) &< x + \alpha h + \varepsilon_1|h| \\
  &= x + \alpha h + \varepsilon h \\
  &< x + \alpha h + \varepsilon'h \\
  &= x + \beta h - (\beta - \alpha - \varepsilon')h \\
  &= x + \beta h - \varepsilon_2|h| \\
  &< \gamma_i(t_2+h).
\end{align*}
That is \(\gamma(t_1+h) \neq \gamma(t_2+h)\).
Therefore \(\gamma\) is not an orbit of a \(\mathbb{R}_+\)-action.

From these results, I lost sight of what is the relation between an autonomous differential equation and a \(\mathbb{R}_+\)-action (or a \(\mathbb{N}\)-action).
I was hoping you could help me find that relation.]]> I'm trying to generalize recurrence relations and autonomous differential equations.
I think that \(\mathbb{R}_+\)-actions are the straightforward continuous version of \(\mathbb{N}\)-actions, so I checked whether an orbit of a \(\mathbb{R}_+\)-action is always a solution (an integral curve) of an autonomous differential equation, and the opposite.

\(\gamma: \mathbb{R} \to \mathbb{R}^n\) is an orbit of a \(\mathbb{R}\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u \in \mathbb{R}\).
\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u > 0\).
\(\gamma\) is a solution of an autonomous differential equation \(\gamma'(t) = F(\gamma(t))\) if and only if \(\gamma(s) = \gamma(t) \implies \gamma'(s) = \gamma'(t)\). (Define \(F(x) := \gamma'(t)\) when \(x = \gamma(t)\). The \(F\) is well-defined on \(\gamma(\mathbb{R})\) because of the hypothesis.)

From the Picard–Lindelöf theorem, if \(F: \mathbb{R}^n \to \mathbb{R}^n\) is Lipschitz continuous, a solution \(\gamma: \mathbb{R} \to \mathbb{R}^n\) of a differential equation \(\gamma'(t) = F(\gamma(t))\) is always an orbit of a \(\mathbb{R}\)-action.
Trivially an orbit of a \(\mathbb{R}\)-action is an orbit of a \(\mathbb{R}_+\)-action.
If an orbit of a \(\mathbb{R}_+\)-action is differentiable,  it's always a solution of an autonomous differential equation. (The proof is described below.)
Therefore let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, then
\[
  \begin{array}{ll}
    & \text{there exists a Lipschitz continuous \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)} \\
    \implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}\)-action} \\
    \implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action} \\
    \implies & \text{there exists \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)}.
  \end{array}
\]
The opposites are not true.

An orbit of a \(\mathbb{R}\)-action that is not a solution of an autonomous differential equation of a Lipschitz continuous function:
Since
\[
  \gamma(t) =
  \begin{cases}
    -t^2 & (t \leq 0) \\
    t^2 & (t \geq 0)
  \end{cases}
\]
is injective, the \(\gamma: \mathbb{R} \to \mathbb{R}^1\) is an orbit of \(\mathbb{R}\)-action.
Assume there exists a Lipschitz continuous \(F: \mathbb{R}^1 \to \mathbb{R}^1\) such that \(\gamma'(t) = F(\gamma(t))\).
Then there exists \(L \in \mathbb{R}\) such that \(|\gamma'(t_1) - \gamma'(t_2)| = |F(\gamma(t_1)) - F(\gamma(t_2))| \leq L|\gamma(t_1) - \gamma(t_2)|\) for all \(t_1, t_2 \in \mathbb{R}\).
Let \(t_1 \neq 0\) and \(t_2 = 0\).
We see \(\gamma'(t) = 2|t|\), so \(2|t_1| \leq L|t_1^2|\). Therefore
\[
  \left|\frac{1}{t_1}\right| = \left|\frac{t_1}{t_1^2}\right| \leq \frac{L}{2}.
\]
\(\left|\frac{1}{t_1}\right| \longrightarrow \infty\) when \(t_1 \longrightarrow 0\), that is a contradiction.

A differentiable orbit of a \(\mathbb{R}_+\)-action that is not an orbit of a \(\mathbb{R}\)-action:
\[
  \gamma: \mathbb{R} \to \mathbb{R}^1, \quad
  \gamma(t) =
  \begin{cases}
    t^2 & (t \leq 0) \\
    0 & (t \geq 0)
  \end{cases}
\]
is an example of it.

A solution of an autonomous differential equation that is not an orbit of a \(\mathbb{R}_+\)-action:
\[
  \gamma: \mathbb{R} \to \mathbb{R}^1, \quad
  \gamma(t) =
  \begin{cases}
    0 & (t \leq 0) \\
    t^2 & (t \geq 0)
  \end{cases}
\]
is a solution of \(\gamma'(t) = F(\gamma(t))\) where
\[
  F: \mathbb{R}^1 \to \mathbb{R}^1, \quad
  F(x) =
  \begin{cases}
    0 & (x \leq 0) \\
    2\sqrt{x} & (x \geq 0),
  \end{cases}
\]
while \(\gamma(-1) = \gamma(0)\) and \(\gamma(-1+1) = 0 \neq 1 = \gamma(0+1)\).

Proof of that a differentiable orbit of a \(\mathbb{R}_+\)-action is a solution of an autonomous differential equation:
We'll prove the contraposition.
Let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, \(\gamma(t) = (\gamma_1(t), \dots, \gamma_n(t))\). (\(\gamma'(t)\) denotes \((\gamma_1'(t), \dots, \gamma_n'(t))\).)
Assume \(\gamma\) is not a a solution of an autonomous differential equation.
Then there are \(t_1, t_2 \in \mathbb{R}\) such that \(\gamma(t_1) = \gamma(t_2)\) and \(\gamma'(t_1) \neq \gamma'(t_2)\).
Therefore \(\gamma'_i(t_1) \neq \gamma'_i(t_2)\) for some \(i\).
Define \(x = \gamma_i(t_1) = \gamma_i(t_2)\), \(\alpha = \gamma_i'(t_1)\), and \(\beta = \gamma_i'(t_2)\). We can assume \(\alpha < \beta\) w.l.o.g.
We know
\[
  \forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall h \in (-\delta_1, \delta_1), \left|\frac{\gamma_i(t_1+h)-x}{h} - \alpha\right| < \varepsilon_1
\]
and
\[
  \forall \varepsilon_2 > 0, \exists \delta_2 > 0, \forall h \in (-\delta_2, \delta_2), \left|\frac{\gamma_i(t_2+h)-x}{h} - \beta\right| < \varepsilon_2.
\]
Thus
\[
  x + \alpha h - \varepsilon_1|h| < \gamma_i(t_1+h) < x + \alpha h + \varepsilon_1|h|,
\]
\[
  x + \beta h - \varepsilon_2|h| < \gamma_i(t_2+h) < x + \beta h + \varepsilon_2|h|.
\]
Since \(\beta - \alpha > 0\) there are \(\varepsilon, \varepsilon' \in \mathbb{R}\) such that \(0 < \varepsilon < \varepsilon' < \beta - \alpha\).
Substitute \(\varepsilon_1 = \varepsilon\) and \(\varepsilon_2 = \beta - \alpha - \varepsilon'\), then for any \(0 < h < \min\{\delta_1, \delta_2\}\),
\begin{align*}
  \gamma_i(t_1+h) &< x + \alpha h + \varepsilon_1|h| \\
  &= x + \alpha h + \varepsilon h \\
  &< x + \alpha h + \varepsilon'h \\
  &= x + \beta h - (\beta - \alpha - \varepsilon')h \\
  &= x + \beta h - \varepsilon_2|h| \\
  &< \gamma_i(t_2+h).
\end{align*}
That is \(\gamma(t_1+h) \neq \gamma(t_2+h)\).
Therefore \(\gamma\) is not an orbit of a \(\mathbb{R}_+\)-action.

From these results, I lost sight of what is the relation between an autonomous differential equation and a \(\mathbb{R}_+\)-action (or a \(\mathbb{N}\)-action).
I was hoping you could help me find that relation.]]> <![CDATA[the fraction iteration approximation of tetration]]> https://tetrationforum.org/showthread.php?tid=1818 Tue, 14 Apr 2026 17:52:40 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1818 we choose a to be 2, b to be -2, c to be ln(a1), d to also be ln(a1)
a1 is the arbitrary base of the logarithm.
when we're done, we use the iteration formula for (ax+b)/(cx+d).
and we get this graph: https://www.desmos.com/calculator/x6malmxjzt]]> we choose a to be 2, b to be -2, c to be ln(a1), d to also be ln(a1)
a1 is the arbitrary base of the logarithm.
when we're done, we use the iteration formula for (ax+b)/(cx+d).
and we get this graph: https://www.desmos.com/calculator/x6malmxjzt]]> <![CDATA[Reviewing recent developments?]]> https://tetrationforum.org/showthread.php?tid=1817 Wed, 31 Dec 2025 06:09:33 +0000 Ember Edison]]> https://tetrationforum.org/showthread.php?tid=1817 Holomorphic Extension of Tetration to Complex Bases and Heights via Schröder's Equation (https://doi.org/10.13140/RG.2.2.10348.48008)? What are its primary conclusions?
How this method compares to the Kneser?]]> Holomorphic Extension of Tetration to Complex Bases and Heights via Schröder's Equation (https://doi.org/10.13140/RG.2.2.10348.48008)? What are its primary conclusions?
How this method compares to the Kneser?]]> <![CDATA[extending normal tetration to real numbers using product tetration]]> https://tetrationforum.org/showthread.php?tid=1816 Fri, 12 Dec 2025 18:49:13 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1816 the base e product tetration has these recursive properties:
\(\text{prodtet}_{e}(x)=\text{prodtet}_{e}(x-1)*e^{\text{prodtet}_{e}(x-1)}\)
\(\text{prodtet}_{e}(x)=\text{W}(\text{prodtet}_{e}(x+1))\)
also, product tetration has this following graph: https://www.desmos.com/calculator/zkjiz8jwh0
if you're wondering how i extended product tetration to real numbers, i used lagrange interpolation on the very far negative side of product tetration and applied recursive formulas.
some example values of product tetration:
\(\text{prodtet}_{e}(-1)=\text{W}(1)=0.56714... \)
\(\text{prodtet}_{e}(0)=1\)
\(\text{prodtet}_{e}(1)=e\)
\(\text{prodtet}_{e}(2)=e^{e+1}\)
and one accepted non-integer value for product tetration:
\(\text{prodtet}_{e}(1/2)=1.5134...\)
and we can express product tetration to different bases by introducing the inverse function of product tetration which will be called the product super logarithm.
the product super logarithm will be denoted as \(\text{prodslog}_{e}(x)\)
the product super logarithm also has some noteworthy recursive properties of its own:
\(\text{prodslog}_{e}(x)=\text{prodslog}_{e}(xe^{x})-1\)
\(\text{prodslog}_{e}(x)=\text{prodslog}_{e}(\text{W}(x))+1\)
the product super logarithm also has a noteworthy graph (sorry if it's slow): https://www.desmos.com/calculator/sxjnv2212j
now for the base change formula of both the product tetration and the product super logarithm:
\( \text{prodtet}_{a}(x)=\frac{\text{prodtet}_{e}(\text{prodslog}_{e}(\text{ln}(a))+x)}{\text{ln}(a)}\)
\(\text{prodslog}_{a}(x)=\text{prodslog}_{e}(x*\text{ln}(a))-\text{prodslog}_{e}(\text{ln}(a))\)
here, i give an example of the base change formulas with base pi: https://www.desmos.com/calculator/kwsqnpe4ef
the proof of the base change formulas is left as an exercise to the reader.
now, back to the main topic of this post.
we can literally extend normal tetration to real numbers by using product tetration because product tetration grows faster than normal tetration.
here's how the process goes:
we extend both the product tetration and the product super logarithm to real numbers using any interpolation technique we consider good. (in this case, i considered the lagrange interpolation good.)
we take the product super logarithm of normal tetration and interpolate the result we get by doing so. (the interpolation still has to be good.)
we finally take the product tetration of the above interpolation to get normal tetration.
here's an example of this process being applied to base 2 tetration: https://www.desmos.com/calculator/r26svg3olm
the numerical accuracy on desmos is bad for product tetration and the product super logarithm.
and desmos can only calculate numbers up to \(2^{1024}\).
i hope there is a calculator that functions like desmos but with greater numerical accuracy and the ability to calculate greater numbers than \(2^{1024}\).]]> the base e product tetration has these recursive properties:
\(\text{prodtet}_{e}(x)=\text{prodtet}_{e}(x-1)*e^{\text{prodtet}_{e}(x-1)}\)
\(\text{prodtet}_{e}(x)=\text{W}(\text{prodtet}_{e}(x+1))\)
also, product tetration has this following graph: https://www.desmos.com/calculator/zkjiz8jwh0
if you're wondering how i extended product tetration to real numbers, i used lagrange interpolation on the very far negative side of product tetration and applied recursive formulas.
some example values of product tetration:
\(\text{prodtet}_{e}(-1)=\text{W}(1)=0.56714... \)
\(\text{prodtet}_{e}(0)=1\)
\(\text{prodtet}_{e}(1)=e\)
\(\text{prodtet}_{e}(2)=e^{e+1}\)
and one accepted non-integer value for product tetration:
\(\text{prodtet}_{e}(1/2)=1.5134...\)
and we can express product tetration to different bases by introducing the inverse function of product tetration which will be called the product super logarithm.
the product super logarithm will be denoted as \(\text{prodslog}_{e}(x)\)
the product super logarithm also has some noteworthy recursive properties of its own:
\(\text{prodslog}_{e}(x)=\text{prodslog}_{e}(xe^{x})-1\)
\(\text{prodslog}_{e}(x)=\text{prodslog}_{e}(\text{W}(x))+1\)
the product super logarithm also has a noteworthy graph (sorry if it's slow): https://www.desmos.com/calculator/sxjnv2212j
now for the base change formula of both the product tetration and the product super logarithm:
\( \text{prodtet}_{a}(x)=\frac{\text{prodtet}_{e}(\text{prodslog}_{e}(\text{ln}(a))+x)}{\text{ln}(a)}\)
\(\text{prodslog}_{a}(x)=\text{prodslog}_{e}(x*\text{ln}(a))-\text{prodslog}_{e}(\text{ln}(a))\)
here, i give an example of the base change formulas with base pi: https://www.desmos.com/calculator/kwsqnpe4ef
the proof of the base change formulas is left as an exercise to the reader.
now, back to the main topic of this post.
we can literally extend normal tetration to real numbers by using product tetration because product tetration grows faster than normal tetration.
here's how the process goes:
we extend both the product tetration and the product super logarithm to real numbers using any interpolation technique we consider good. (in this case, i considered the lagrange interpolation good.)
we take the product super logarithm of normal tetration and interpolate the result we get by doing so. (the interpolation still has to be good.)
we finally take the product tetration of the above interpolation to get normal tetration.
here's an example of this process being applied to base 2 tetration: https://www.desmos.com/calculator/r26svg3olm
the numerical accuracy on desmos is bad for product tetration and the product super logarithm.
and desmos can only calculate numbers up to \(2^{1024}\).
i hope there is a calculator that functions like desmos but with greater numerical accuracy and the ability to calculate greater numbers than \(2^{1024}\).]]> <![CDATA[Confirmed: the constancy of the congruence speed holds in each squarefree radix]]> https://tetrationforum.org/showthread.php?tid=1815 Mon, 01 Dec 2025 19:28:32 +0000 marcokrt]]> https://tetrationforum.org/showthread.php?tid=1815 I would like to share a very compact one-page note that summarizes six fundamental equations describing the constant congruence speed of integer tetration bases (greater than 1 and not a multiple of the selected radix) in arbitrary numeral systems \(r > 1\).
These few equations condense the key results established in our referenced recent works, including the complete formulas for prime radices, the explicit radix-6 and radix-10 cases, and a couple of elegant identities valid for all squarefree \(r > 2\).

In particular, I am especially fond of identity (4). It highlights the structural role of the \(r\)-adic fixed point \(1_r\): it considers 5 different parameters on the left-hand side, including the height of the power tower, which only needs to be greater than 1, and for \(k = 0\) (by Mihăilescu's theorem) it returns only perfect powers of exact order \(c\) whose radix-\(r\) constant congruence speed is exactly \(t\). Furthermore, it includes hyper-1 as sum/difference, hyper-2 as product, hyper-3 as exponentiation, and finally the function \(V\) arises from tetration (i.e., hyper-4).

I am attaching these formulas as a figure; the one-page note is also available on Zenodo: Six fundamental equations for the constant congruence speed in any numeral system.

Comments, questions, and suggestions are of course welcome... it may also be interesting to discuss possible extensions or implications for related r-adic fixed-point structures or iterated-exponentiation phenomena (I have also developed a very rough key-exchange idea based on solving fundamental equations of the form \(y^t = y\) for sufficiently large odd values of \(t := t\)(\(r\)) in the ring of \(r\)-adic integers \(\mathbb{Z}_r\)).

.jpg   Six fundamental equations (reduced).jpg (Size: 228.37 KB / Downloads: 228) ]]> I would like to share a very compact one-page note that summarizes six fundamental equations describing the constant congruence speed of integer tetration bases (greater than 1 and not a multiple of the selected radix) in arbitrary numeral systems \(r > 1\).
These few equations condense the key results established in our referenced recent works, including the complete formulas for prime radices, the explicit radix-6 and radix-10 cases, and a couple of elegant identities valid for all squarefree \(r > 2\).

In particular, I am especially fond of identity (4). It highlights the structural role of the \(r\)-adic fixed point \(1_r\): it considers 5 different parameters on the left-hand side, including the height of the power tower, which only needs to be greater than 1, and for \(k = 0\) (by Mihăilescu's theorem) it returns only perfect powers of exact order \(c\) whose radix-\(r\) constant congruence speed is exactly \(t\). Furthermore, it includes hyper-1 as sum/difference, hyper-2 as product, hyper-3 as exponentiation, and finally the function \(V\) arises from tetration (i.e., hyper-4).

I am attaching these formulas as a figure; the one-page note is also available on Zenodo: Six fundamental equations for the constant congruence speed in any numeral system.

Comments, questions, and suggestions are of course welcome... it may also be interesting to discuss possible extensions or implications for related r-adic fixed-point structures or iterated-exponentiation phenomena (I have also developed a very rough key-exchange idea based on solving fundamental equations of the form \(y^t = y\) for sufficiently large odd values of \(t := t\)(\(r\)) in the ring of \(r\)-adic integers \(\mathbb{Z}_r\)).

.jpg   Six fundamental equations (reduced).jpg (Size: 228.37 KB / Downloads: 228) ]]> <![CDATA[what we should call the hyperoperations in different languages]]> https://tetrationforum.org/showthread.php?tid=1814 Sat, 15 Nov 2025 15:12:15 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1814 romanian: tetrație, pentație, hexație, etc.
icelandic: 4 rekstur, 5 rekstur, 6 rekstur, etc.
note: in icelandic, rekstur means operation.
german, french: same as english.
second note: we already have the russian, turkish, spanish, and italian names for hyperoperations.
mongolian: 4 үйл, 5 үйл, 6 үйл, etc.
third note: in mongolian, үйл means operation.
hungarian: 4 művelet, 5 művelet, 6 művelet, etc.
fourth note: in hungarian, művelet means operation.
you can add to the list if you know a language other than the ones mentioned here.]]> romanian: tetrație, pentație, hexație, etc.
icelandic: 4 rekstur, 5 rekstur, 6 rekstur, etc.
note: in icelandic, rekstur means operation.
german, french: same as english.
second note: we already have the russian, turkish, spanish, and italian names for hyperoperations.
mongolian: 4 үйл, 5 үйл, 6 үйл, etc.
third note: in mongolian, үйл means operation.
hungarian: 4 művelet, 5 művelet, 6 művelet, etc.
fourth note: in hungarian, művelet means operation.
you can add to the list if you know a language other than the ones mentioned here.]]> <![CDATA[Tetration with complex bases]]> https://tetrationforum.org/showthread.php?tid=1813 Thu, 13 Nov 2025 09:33:54 +0000 TetrationSheep]]> https://tetrationforum.org/showthread.php?tid=1813 <![CDATA[Orbit-like maps on linearly ordered groups]]> https://tetrationforum.org/showthread.php?tid=1812 Sat, 08 Nov 2025 07:47:16 +0000 Natsugou]]> https://tetrationforum.org/showthread.php?tid=1812 For a map \(a: \mathbb{Z} \to X, n \mapsto a_n\), there exists \(f: X \to X\) such that \(a_{n+1} = f(a_n)\) if and only if \(a_m = a_n \implies a_{m+k} = a_{n+k}\) for any positive integer \(k\).
When \(a: \mathbb{Z} \to X\) satisfies the above condition, the periods of \(a\), \(P(a) = \{m-n \mid a_m = a_n\}\) is a subgroup of \(\mathbb{Z}\).

To generalize to arbitrary linearly ordered groups, I will introduce some definitions.
For a map \(a\) defined on a linearly ordered group \(G\),
\(a\) is peep if \(a(g) = a(h) \implies a(fg) = a(fh)\) for all \(f > 1\).
\(a\) is bib if \(a(g) = a(h) \implies a(gf) = a(hf)\) for all \(f > 1\).
\(a\) is peep bib if \(a\) is peep and bib.
I define periods of \(a\) as
\begin{align}
  P(a) &= \{g^{-1}h \mid a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g, h \in G, p = g^{-1}h\: \text{and}\: a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g \in G, a(g) = a(gp)\}.
\end{align}
A subgroup \(H\) of \(G\) is said to be fee if \(g^{-1}Hg \subset H\) for all \(g > 1\).

Let \(G\) be a bi-ordered group and \(a\) be a map defined on \(G\). (Edit: \(G\) need not to be bi-ordered, just need to be linearly ordered.) Then
\begin{equation}
  \begin{array}{ccccc}
  & & a\: \text{is}\: \textit{peep bib} & \implies & a\: \text{is}\: \textit{peep} \\
  & & \Downarrow & & \Downarrow \\
  P(a)\: \text{is normal} & \implies & P(a)\: \text{is}\: \textit{fee} & \implies & P(a)\: \text{is a subgroup} \\
  \end{array}
\end{equation}
and the converses of these are not true.

For example, let \(F\) be the free group generated by \(a, b\) and bi-ordred in the way of Definition 3 of https://doi.org/10.4153/CJM-2003-034-2
Then the subgroup \(\langle a \rangle\) is not fee.
The fee subgroup \(\langle b^{F_+} \rangle\) is not normal (shown in the attached pdf
.pdf   momo (4).pdf (Size: 223.12 KB / Downloads: 155) ).
We define a map \(a\) on \(\mathbb{Z}\) as
\begin{equation}
  \dots, -7, -6, -5, 22, -3, 22, -1, 0, 1, 0, 1, 0, 1, 0, 1, \dots
\end{equation}
Then \(a\) is neither peep nor bib, and \(P(a) = 2\mathbb{Z}\).]]> For a map \(a: \mathbb{Z} \to X, n \mapsto a_n\), there exists \(f: X \to X\) such that \(a_{n+1} = f(a_n)\) if and only if \(a_m = a_n \implies a_{m+k} = a_{n+k}\) for any positive integer \(k\).
When \(a: \mathbb{Z} \to X\) satisfies the above condition, the periods of \(a\), \(P(a) = \{m-n \mid a_m = a_n\}\) is a subgroup of \(\mathbb{Z}\).

To generalize to arbitrary linearly ordered groups, I will introduce some definitions.
For a map \(a\) defined on a linearly ordered group \(G\),
\(a\) is peep if \(a(g) = a(h) \implies a(fg) = a(fh)\) for all \(f > 1\).
\(a\) is bib if \(a(g) = a(h) \implies a(gf) = a(hf)\) for all \(f > 1\).
\(a\) is peep bib if \(a\) is peep and bib.
I define periods of \(a\) as
\begin{align}
  P(a) &= \{g^{-1}h \mid a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g, h \in G, p = g^{-1}h\: \text{and}\: a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g \in G, a(g) = a(gp)\}.
\end{align}
A subgroup \(H\) of \(G\) is said to be fee if \(g^{-1}Hg \subset H\) for all \(g > 1\).

Let \(G\) be a bi-ordered group and \(a\) be a map defined on \(G\). (Edit: \(G\) need not to be bi-ordered, just need to be linearly ordered.) Then
\begin{equation}
  \begin{array}{ccccc}
  & & a\: \text{is}\: \textit{peep bib} & \implies & a\: \text{is}\: \textit{peep} \\
  & & \Downarrow & & \Downarrow \\
  P(a)\: \text{is normal} & \implies & P(a)\: \text{is}\: \textit{fee} & \implies & P(a)\: \text{is a subgroup} \\
  \end{array}
\end{equation}
and the converses of these are not true.

For example, let \(F\) be the free group generated by \(a, b\) and bi-ordred in the way of Definition 3 of https://doi.org/10.4153/CJM-2003-034-2
Then the subgroup \(\langle a \rangle\) is not fee.
The fee subgroup \(\langle b^{F_+} \rangle\) is not normal (shown in the attached pdf
.pdf   momo (4).pdf (Size: 223.12 KB / Downloads: 155) ).
We define a map \(a\) on \(\mathbb{Z}\) as
\begin{equation}
  \dots, -7, -6, -5, 22, -3, 22, -1, 0, 1, 0, 1, 0, 1, 0, 1, \dots
\end{equation}
Then \(a\) is neither peep nor bib, and \(P(a) = 2\mathbb{Z}\).]]> <![CDATA[i have a seemingly reasonable definition for tetration]]> https://tetrationforum.org/showthread.php?tid=1810 Sun, 19 Oct 2025 15:00:11 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1810 1. i construct the xth super root of y by lagrange interpolation.
2. i evaluate my constructed super root function at a non-integer x value.
3. i finally construct my tetration function by using the y value i got for my super root function.
more details are given in this desmos graph: https://www.desmos.com/calculator/ooxd3v6twb
as can be seen in the desmos graph, i constructed the xth super root of 2 and i chose 3.5 as my x value.
that's how i constructed the 3.5th super root of 2 tetrated to x.
if you want to know the value for the 3.5th super root of 2 using my method and kneser's, here it is:
my method: \( \sqrt[3.5]{2}_{s}=1.45849906738\)
kneser's method:  \( \sqrt[3.5]{2}_{s}=1.4584946676580992\)
my method and kneser's method give slightly different results. that might be because the lagrange interpolation from my method does not approximate the super root function as well as kneser's method. but maybe the approximation problem can be solved using limits.
here's what i mean: we take the limit of the xth super root as x goes to infinity.
we interpolate the function using the limit values and we apply recursive formulas to construct the xth super root.
it doesn't seem like it's easy to find recursive formulas for the xth super root.
like i said in the title of this thread, my tetration method SEEMS reasonable. but i don't yet know if it IS reasonable.
maybe someone finds a quick algorithm in desmos to compute super roots. the algorithm i found required a lot of iterations of newton's method. too many iterations of newton's method would be bad for my computer.
and another point i would like to make: it seems like i can continue on a similar path for pentation, hexation, etc.
and by continuing on a similar path, i mean adapting my tetration method for higher hyperoperations.]]> 1. i construct the xth super root of y by lagrange interpolation.
2. i evaluate my constructed super root function at a non-integer x value.
3. i finally construct my tetration function by using the y value i got for my super root function.
more details are given in this desmos graph: https://www.desmos.com/calculator/ooxd3v6twb
as can be seen in the desmos graph, i constructed the xth super root of 2 and i chose 3.5 as my x value.
that's how i constructed the 3.5th super root of 2 tetrated to x.
if you want to know the value for the 3.5th super root of 2 using my method and kneser's, here it is:
my method: \( \sqrt[3.5]{2}_{s}=1.45849906738\)
kneser's method:  \( \sqrt[3.5]{2}_{s}=1.4584946676580992\)
my method and kneser's method give slightly different results. that might be because the lagrange interpolation from my method does not approximate the super root function as well as kneser's method. but maybe the approximation problem can be solved using limits.
here's what i mean: we take the limit of the xth super root as x goes to infinity.
we interpolate the function using the limit values and we apply recursive formulas to construct the xth super root.
it doesn't seem like it's easy to find recursive formulas for the xth super root.
like i said in the title of this thread, my tetration method SEEMS reasonable. but i don't yet know if it IS reasonable.
maybe someone finds a quick algorithm in desmos to compute super roots. the algorithm i found required a lot of iterations of newton's method. too many iterations of newton's method would be bad for my computer.
and another point i would like to make: it seems like i can continue on a similar path for pentation, hexation, etc.
and by continuing on a similar path, i mean adapting my tetration method for higher hyperoperations.]]> <![CDATA[my proposed extension of the fast growing hierarchy to real numbers]]> https://tetrationforum.org/showthread.php?tid=1809 Sun, 28 Sep 2025 18:15:55 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1809 https://www.desmos.com/calculator/i9ue4cyvan
i took the superfunction of \(f_{2}(x)\) on the integer points and used lagrange interpolation on the left side of the superfunction.
and that's how i got \(f_{3}(x)\).
and you can go through the same process to get \(f_{4}(x)\), \(f_{5}(x)\), \(f_{6}(x)\), and so on...
but i don't even know how \(f_{\omega}(x)\) can be extended.
we first have to define the fast growing hierarchy for non-integer subscripts.
\(f_{n}(1)=2\) for all n. and we can choose to keep it true for non-integer values of n.
the only catch is that we don't yet have a full understanding of the fast growing hierarchy with real numbers.
we can't just start taking derivatives, plugging in values, or anything like that.
we have to actually understand the fast growing hierarchy.
if we understand the fast growing hierarchy well enough, we can even extend it to complex numbers.]]> https://www.desmos.com/calculator/i9ue4cyvan
i took the superfunction of \(f_{2}(x)\) on the integer points and used lagrange interpolation on the left side of the superfunction.
and that's how i got \(f_{3}(x)\).
and you can go through the same process to get \(f_{4}(x)\), \(f_{5}(x)\), \(f_{6}(x)\), and so on...
but i don't even know how \(f_{\omega}(x)\) can be extended.
we first have to define the fast growing hierarchy for non-integer subscripts.
\(f_{n}(1)=2\) for all n. and we can choose to keep it true for non-integer values of n.
the only catch is that we don't yet have a full understanding of the fast growing hierarchy with real numbers.
we can't just start taking derivatives, plugging in values, or anything like that.
we have to actually understand the fast growing hierarchy.
if we understand the fast growing hierarchy well enough, we can even extend it to complex numbers.]]> <![CDATA[product tetration]]> https://tetrationforum.org/showthread.php?tid=1808 Wed, 24 Sep 2025 11:47:13 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1808 it uses the lambert w function (also known as the product logarithm) to go in the negative direction.
product tetration requires three variables and is denoted as: \(\text{prodtet}(a,b,c)\)
product tetration can be interpolated through lagrange polynomials for extension to real numbers.
and it has this recursive property:
\(\text{prodtet}(a,b,c)= f(\text{prodtet}(a,b,c-1))\)
and this base case:
\(\text{prodtet}(a,b,0)= b\)
more details are given on this desmos graph: https://www.desmos.com/calculator/cnvdx5ce1r]]> it uses the lambert w function (also known as the product logarithm) to go in the negative direction.
product tetration requires three variables and is denoted as: \(\text{prodtet}(a,b,c)\)
product tetration can be interpolated through lagrange polynomials for extension to real numbers.
and it has this recursive property:
\(\text{prodtet}(a,b,c)= f(\text{prodtet}(a,b,c-1))\)
and this base case:
\(\text{prodtet}(a,b,0)= b\)
more details are given on this desmos graph: https://www.desmos.com/calculator/cnvdx5ce1r]]> <![CDATA[self penta root and infinite hexation]]> https://tetrationforum.org/showthread.php?tid=1807 Sat, 30 Aug 2025 21:07:24 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1807 so i decided that i would make this graph of the self penta root and infinite hexation: https://www.desmos.com/calculator/j99fxd2uzw?lang=en
i used some of my graphs from my community thread and this holomorphic tetration calculator: https://www.desmos.com/calculator/mvdkr1rp8i?lang=ru
and i used lagrange interpolation on the points that i could calculate for the self penta root.
the holomorphic tetration calculator took painfully long so i could only plot a few points. Sad
i hope someone finds a faster way to implement tetration, pentation, etc. in desmos.]]> so i decided that i would make this graph of the self penta root and infinite hexation: https://www.desmos.com/calculator/j99fxd2uzw?lang=en
i used some of my graphs from my community thread and this holomorphic tetration calculator: https://www.desmos.com/calculator/mvdkr1rp8i?lang=ru
and i used lagrange interpolation on the points that i could calculate for the self penta root.
the holomorphic tetration calculator took painfully long so i could only plot a few points. Sad
i hope someone finds a faster way to implement tetration, pentation, etc. in desmos.]]> <![CDATA[[Question] Classifying dynamical system by connected components]]> https://tetrationforum.org/showthread.php?tid=1802 Fri, 06 Jun 2025 00:39:10 +0000 MphLee]]> https://tetrationforum.org/showthread.php?tid=1802
Using this framework, we can approach discrete dynamics in a more visual and geometric manner. We can classify systems in a way very reminiscent of how we can classify vector spaces or other geometric objects.

Connected dynamical systems 
We can ask if \(X\) is connected. This happens if given two states of the system \(x,y\) there exists a third state \(z\) that connects them in the dynamics, i.e., such that \(f^n(x)=z=f^m(y)\). Connected dynamical systems can be visualized as made by only a single piece. Some examples of connected systems are:
  • \( \bar { \mathbb N}=(\mathbb N, S) \) the naturals with successor;
  • \( \bar { \mathbb Z}=(\mathbb Z, S) \) the integers with successor;
  • \( \bar { \mathbb Z}_m=(\mathbb Z_m, S) \) this is the \(m\) element set with successor modulo \(m\), i.e., \(S(n):=n+1 \,{\rm mod} \,m\);
  • \( \bar { \Omega}=(\mathbb N, P) \) this is the natural number with cut-off predecessor, i.e., \(P(n)={\rm max}(0,n-1)\);
  • for \((p,h)\in \mathbb N^+\times \mathbb N\) define \( \bar { \mathbb N}_{(p,h)}=(\mathbb N, S) \) as \(p+h\) element set \(\{0,1,...,h, h+1,...,h+p-1\}\) with dynamic \(S(n)=n+1\) and \(S(h+p-1)=h\);
  • \(K=(\mathbb N, 0\cdot)\) is the constant function with \(0\) as the only fixed point where every \(n\) is mapped to \(0\);
  • \((\mathbb N[x],\frac{\rm d}{{\rm d}x})\) where the formal polynomial with natural coefficients \((a_0,a_1,a_2,...,a_d)\) is mapped to its formal derivative \((a_1,2a_2,3a_3,...,na_n)\). This dynamical system has a single fixed point.

Non-connected dynamical systems 
A system that is non-connected is the union of a finite or an infinite number of connected dynamical systems. Denote with \(\pi_0 X\) the set of connected components of \(X\); it is a partition of \(X\) into equivalence classes by the relation of "connected states." If \(\|\pi_0X\|=1\), then we have a connected dynamical system. We can see the ordinal \(\beta_0(X)=\|\pi_0X\|\) as a kind of Betti number of the dynamical system. This means that for each dynamical system \(X=(X,f)\), we can find a unique decomposition 
\[\boxed{X=\bigsqcup_{\lambda\in \pi_0X} X_\lambda}\]
as a sum of connected dynamical systems \(X_\lambda\). We have the following decompositions of non-connected dynamical systems:
  • Let \(\bar{\mathbb R}=(\mathbb R,S)\) be the real numbers and the successor function.
  • Let \( h_{1,b}=(\mathbb N, h_{1,b}) \) be the naturals with addition by \(b\), i.e., \(h_{1,b}(n)=b+n\);
  • Let \( h_{2,b}=(\mathbb N, h_{2,b}) \) be the naturals with multiplication by \(b\), i.e., \(h_{2,b}(n)=b\cdot n\);
  • Let \( {\rm Collatz}=(\mathbb N, f) \) be the naturals with the Collatz map.

Proposition: 
i) First of all, the Collatz dynamic is connected, \(\|\pi_0 ({\rm Collatz})\|=1\) iff the Collatz's conjecture is true.
ii) For the real numbers \(\pi_0\bar{\mathbb R}\cong [0,1)\) and since each real number can be decomposed in integer and fractional part \(r,n\) because \(x=r+n\) for \(r\in[0,1)\) we decompose the dynamics as \[\bar{\mathbb R}\cong [0,1)\bar{\mathbb N}\]
iii) Then \(\|\pi_0 (h_{1,b})\|=b\) if \(b\neq 0\) \[h_{1,b}\cong b \bar { \mathbb N}\] each state is decomposed uniquely as \((r,h)\) where \(n=r+bh\) with \(0\leq r < b\) (this is just the Euclidean division). 
iv) And \(\|\pi_0 (h_{1,b})\|=\aleph_0\) if \(b=0\) since we have the dynamic of the identity where each state is a fixed point and \(h_{1,0}\cong \aleph_0 \bar { \mathbb Z}_1\). 
v) For rank 2, i.e., multiplication, we have \(h_{2,0}=K\) thus \( \|\pi_0 (h_{2,0})\|=1\); \(h_{2,1}=h_{1,0}\) is the identity and \(\|\pi_0 (h_{2,b})\|=\aleph_0\) has infinite connected components if \(b\geq 2\). In the infinite case, we have the decomposition \[h_{2,b}\cong 1 \sqcup \mathbb N[x] \bar { \mathbb N}\] this decomposes each number \(n\) as the pair \((r,h)\) where \(b\) does not divide \(r\) and such that \(n=r\cdot b^h\).

Motivation
Why this matters? Because if let \({\rm hom}(X,Y)\) be the set of dynamical system maps \( \phi : (X,f)\to(Y,g) \), or equivarian maps with \( \phi(f(x))=g(\phi(x)) \) then
  • \({\rm hom}(X,\bar { \mathbb N})\) is the set of positive integer valued abel functions of \(X\);
  • \({\rm hom}(X,\bar { \mathbb R})\) is the set of real valued abel functions of \(X\);
  • \({\rm hom}(X,h_{2,b})\) is the set of positive integer valued schroeder functions of \(X\);
  • \({\rm hom}(\bar { \mathbb N},X) \) contains the orbits of \(X\);
  • \({\rm hom}(\bar { \mathbb Z}_1,X) \) contains the all fixed points of \(X\), also called equilibria;
  • \({\rm hom}(\bar { \mathbb Z}_n,X) \) contains the all n-periodic points of \(X\);
  • \({\rm hom}(\bar { \mathbb R},X) \) contains the all the possible real superfunctions of \(X\), continuous, analytic and disconitnuous;
  • \({\rm hom}(\bar { \mathbb R},h_{3,b}) \) contains the all the possible real superfunctions of exponentiation, also the tetration;

If we know the decomposition in connected components of \(X=\bigsqcup_{\lambda\in \pi_0X} X_{\lambda}\) we can derive a parametrization of the hom-sets since
\[\boxed{ {\rm hom}(\bigsqcup_{\lambda\in \pi_0X} X_{\lambda},Y)\cong \prod_{\lambda\in \pi_0X}{\rm hom}(X_{\lambda},Y) }\]


Quote:Question 
What is known about \(\pi_0 f\) for \(f\) being the logarithm or the exponentiation over the real numbers? Over the complex numbers?
]]>
Using this framework, we can approach discrete dynamics in a more visual and geometric manner. We can classify systems in a way very reminiscent of how we can classify vector spaces or other geometric objects.

Connected dynamical systems 
We can ask if \(X\) is connected. This happens if given two states of the system \(x,y\) there exists a third state \(z\) that connects them in the dynamics, i.e., such that \(f^n(x)=z=f^m(y)\). Connected dynamical systems can be visualized as made by only a single piece. Some examples of connected systems are:
  • \( \bar { \mathbb N}=(\mathbb N, S) \) the naturals with successor;
  • \( \bar { \mathbb Z}=(\mathbb Z, S) \) the integers with successor;
  • \( \bar { \mathbb Z}_m=(\mathbb Z_m, S) \) this is the \(m\) element set with successor modulo \(m\), i.e., \(S(n):=n+1 \,{\rm mod} \,m\);
  • \( \bar { \Omega}=(\mathbb N, P) \) this is the natural number with cut-off predecessor, i.e., \(P(n)={\rm max}(0,n-1)\);
  • for \((p,h)\in \mathbb N^+\times \mathbb N\) define \( \bar { \mathbb N}_{(p,h)}=(\mathbb N, S) \) as \(p+h\) element set \(\{0,1,...,h, h+1,...,h+p-1\}\) with dynamic \(S(n)=n+1\) and \(S(h+p-1)=h\);
  • \(K=(\mathbb N, 0\cdot)\) is the constant function with \(0\) as the only fixed point where every \(n\) is mapped to \(0\);
  • \((\mathbb N[x],\frac{\rm d}{{\rm d}x})\) where the formal polynomial with natural coefficients \((a_0,a_1,a_2,...,a_d)\) is mapped to its formal derivative \((a_1,2a_2,3a_3,...,na_n)\). This dynamical system has a single fixed point.

Non-connected dynamical systems 
A system that is non-connected is the union of a finite or an infinite number of connected dynamical systems. Denote with \(\pi_0 X\) the set of connected components of \(X\); it is a partition of \(X\) into equivalence classes by the relation of "connected states." If \(\|\pi_0X\|=1\), then we have a connected dynamical system. We can see the ordinal \(\beta_0(X)=\|\pi_0X\|\) as a kind of Betti number of the dynamical system. This means that for each dynamical system \(X=(X,f)\), we can find a unique decomposition 
\[\boxed{X=\bigsqcup_{\lambda\in \pi_0X} X_\lambda}\]
as a sum of connected dynamical systems \(X_\lambda\). We have the following decompositions of non-connected dynamical systems:
  • Let \(\bar{\mathbb R}=(\mathbb R,S)\) be the real numbers and the successor function.
  • Let \( h_{1,b}=(\mathbb N, h_{1,b}) \) be the naturals with addition by \(b\), i.e., \(h_{1,b}(n)=b+n\);
  • Let \( h_{2,b}=(\mathbb N, h_{2,b}) \) be the naturals with multiplication by \(b\), i.e., \(h_{2,b}(n)=b\cdot n\);
  • Let \( {\rm Collatz}=(\mathbb N, f) \) be the naturals with the Collatz map.

Proposition: 
i) First of all, the Collatz dynamic is connected, \(\|\pi_0 ({\rm Collatz})\|=1\) iff the Collatz's conjecture is true.
ii) For the real numbers \(\pi_0\bar{\mathbb R}\cong [0,1)\) and since each real number can be decomposed in integer and fractional part \(r,n\) because \(x=r+n\) for \(r\in[0,1)\) we decompose the dynamics as \[\bar{\mathbb R}\cong [0,1)\bar{\mathbb N}\]
iii) Then \(\|\pi_0 (h_{1,b})\|=b\) if \(b\neq 0\) \[h_{1,b}\cong b \bar { \mathbb N}\] each state is decomposed uniquely as \((r,h)\) where \(n=r+bh\) with \(0\leq r < b\) (this is just the Euclidean division). 
iv) And \(\|\pi_0 (h_{1,b})\|=\aleph_0\) if \(b=0\) since we have the dynamic of the identity where each state is a fixed point and \(h_{1,0}\cong \aleph_0 \bar { \mathbb Z}_1\). 
v) For rank 2, i.e., multiplication, we have \(h_{2,0}=K\) thus \( \|\pi_0 (h_{2,0})\|=1\); \(h_{2,1}=h_{1,0}\) is the identity and \(\|\pi_0 (h_{2,b})\|=\aleph_0\) has infinite connected components if \(b\geq 2\). In the infinite case, we have the decomposition \[h_{2,b}\cong 1 \sqcup \mathbb N[x] \bar { \mathbb N}\] this decomposes each number \(n\) as the pair \((r,h)\) where \(b\) does not divide \(r\) and such that \(n=r\cdot b^h\).

Motivation
Why this matters? Because if let \({\rm hom}(X,Y)\) be the set of dynamical system maps \( \phi : (X,f)\to(Y,g) \), or equivarian maps with \( \phi(f(x))=g(\phi(x)) \) then
  • \({\rm hom}(X,\bar { \mathbb N})\) is the set of positive integer valued abel functions of \(X\);
  • \({\rm hom}(X,\bar { \mathbb R})\) is the set of real valued abel functions of \(X\);
  • \({\rm hom}(X,h_{2,b})\) is the set of positive integer valued schroeder functions of \(X\);
  • \({\rm hom}(\bar { \mathbb N},X) \) contains the orbits of \(X\);
  • \({\rm hom}(\bar { \mathbb Z}_1,X) \) contains the all fixed points of \(X\), also called equilibria;
  • \({\rm hom}(\bar { \mathbb Z}_n,X) \) contains the all n-periodic points of \(X\);
  • \({\rm hom}(\bar { \mathbb R},X) \) contains the all the possible real superfunctions of \(X\), continuous, analytic and disconitnuous;
  • \({\rm hom}(\bar { \mathbb R},h_{3,b}) \) contains the all the possible real superfunctions of exponentiation, also the tetration;

If we know the decomposition in connected components of \(X=\bigsqcup_{\lambda\in \pi_0X} X_{\lambda}\) we can derive a parametrization of the hom-sets since
\[\boxed{ {\rm hom}(\bigsqcup_{\lambda\in \pi_0X} X_{\lambda},Y)\cong \prod_{\lambda\in \pi_0X}{\rm hom}(X_{\lambda},Y) }\]


Quote:Question 
What is known about \(\pi_0 f\) for \(f\) being the logarithm or the exponentiation over the real numbers? Over the complex numbers?
]]> <![CDATA[Tetration using the integral]]> https://tetrationforum.org/showthread.php?tid=1800 Tue, 27 May 2025 22:27:04 +0000 Pentalogue]]> https://tetrationforum.org/showthread.php?tid=1800 I wondered what definite integral could be used to describe the tetration so that the function passes through all points of the tetration results with integer tetration exponents. This question arose because it is currently known that the gamma function is a complete analytic extension (one of the variants) of the factorial, and that it was not previously known what the solutions were for the factorial, which would be taken from a non-integer number. Only then did we calculate the integral that realizes the factorial function as a full analytic continuation, where the curve clearly passes through all points of factorials of integers, starting from zero factorial. The factorial is an iterative function because it is based on recursion, so the gamma function is also an iterative function.

.jpg   Screenshot_2025_0510_212419.jpg (Size: 58.61 KB / Downloads: 587) ]]> I wondered what definite integral could be used to describe the tetration so that the function passes through all points of the tetration results with integer tetration exponents. This question arose because it is currently known that the gamma function is a complete analytic extension (one of the variants) of the factorial, and that it was not previously known what the solutions were for the factorial, which would be taken from a non-integer number. Only then did we calculate the integral that realizes the factorial function as a full analytic continuation, where the curve clearly passes through all points of factorials of integers, starting from zero factorial. The factorial is an iterative function because it is based on recursion, so the gamma function is also an iterative function.

.jpg   Screenshot_2025_0510_212419.jpg (Size: 58.61 KB / Downloads: 587) ]]> <![CDATA[Not about tetration but what do you think about this paper?]]> https://tetrationforum.org/showthread.php?tid=1799 Fri, 11 Apr 2025 03:27:28 +0000 Ansus]]> https://tetrationforum.org/showthread.php?tid=1799
so, here I invite to discuss this new paper of mine:

https://arxiv.org/abs/2411.00296]]>
so, here I invite to discuss this new paper of mine:

https://arxiv.org/abs/2411.00296]]>