https://tetrationforum.org/ Tue, 12 May 2026 10:50:19 +0000 MyBB 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.]]> 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]]> 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?]]> 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 \(2^{1024}\).]]> 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 \(\mathbb{Z}_r\)).

  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 \(\mathbb{Z}_r\)).

  Six fundamental equations (reduced).jpg (Size: 228.37 KB / Downloads: 228) ]]> 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.]]> https://tetrationforum.org/showthread.php?tid=1813 Thu, 13 Nov 2025 09:33:54 +0000 TetrationSheep]]> https://tetrationforum.org/showthread.php?tid=1813 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
  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
  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 \(P(a) = 2\mathbb{Z}\).]]> https://tetrationforum.org/showthread.php?tid=1811 Mon, 20 Oct 2025 15:28:21 +0000 Gottfried]]> https://tetrationforum.org/showthread.php?tid=1811
 it is about 6 weeks, that some correspondent from USA told me he cannot access the files on my website. It seemed so as well from Scotland - but the access from Germany seems to work (I can access everything by ftp:// and even by https:// ).
Now the admin of our small network https://helms-net.de tells me he had tried to reverse some changes he had made in spring, and I would like to check, whether international folks can now access my work again.

Could someone please try to access my math-homepage  

        https://go.helms-net.de/math/tetdocs/index.htm

and if that is possible, then to download just a single *.pdf-file as test?

Please tell me the result here, or even better, mail me at

     mailto:gottfried.helms@t-online.de

(possibly include the error-message as well)


Thank you in advance -

Gottfried.]]>
 it is about 6 weeks, that some correspondent from USA told me he cannot access the files on my website. It seemed so as well from Scotland - but the access from Germany seems to work (I can access everything by ftp:// and even by https:// ).
Now the admin of our small network https://helms-net.de tells me he had tried to reverse some changes he had made in spring, and I would like to check, whether international folks can now access my work again.

Could someone please try to access my math-homepage  

        https://go.helms-net.de/math/tetdocs/index.htm

and if that is possible, then to download just a single *.pdf-file as test?

Please tell me the result here, or even better, mail me at

     mailto:gottfried.helms@t-online.de

(possibly include the error-message as well)


Thank you in advance -

Gottfried.]]> 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.]]> 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.]]> 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]]> https://tetrationforum.org/showthread.php?tid=1806 Sat, 30 Aug 2025 22:05:54 +0000 Rayanso]]> https://tetrationforum.org/showthread.php?tid=1806
On this forum I discovered GFR's post and I looked at the paper, and I also used the wayback machine to download an old paper from 1996 on hyper-operations from KAR's website.
If you are not familiar with Zeration and Deltation I suggest that you read the file that GFR attached in his post.

One question I was asking myself was what 0.5 * (△0) would be equal to, and if it could be reduced to a single number. In KAR's paper, this is resolved by introducing a new unit "j", like the imaginary unit for (-1)^(0.5). We then have 2*j =  △0. The operation 0.5 * (△0) also has 2 solutions: △j and j.

We still don't know what 0.35801 * (△0), or with any other arbitrary coefficient, would equal to. I made a formula for any real number as a coefficient for what this would be equal to.

I came up with x * (△0) = j*2*(2n+1)*x with n being an integer. I verified this formula with (-1)^a = b^(△0 * a)

Verification: for (-1)^(1/2) we have b^(△0 * 1/2) = {b^j ; b^(△j)}, so we must have b^j = i
b is a real number that is not zero.

For (-1)^(1/3) we know that there will be 3 solutions, therefore, △0 * 1/3 will have 3 solutions. These are: △0 ; 2/3 * j ; -2/3 * j. All other solutions can be reduced to one of these 3.

We also have an interesting rule that completes one that already exists:

a^0 = 1
a^(△0) = -1
a^j = i
a^(△j) = -i

This makes sense because the absolute value is 1 in every case, and the exponent is linked in some way to the number 0. 
If we have b = a*(△0), for any real a. c^b will always have an absolute value of 1. 

I also discovered the rule △j = -j, which I think KAR has not mentioned in his paper.

Proof:

△j + △j = 2*△j = △0
△j + (-j) = △0 + j - j = △0
(-j) + (-j) = -(2*j) = -△0 = △0

This is as far as I went and there are many gaps in the algebraic structure, KAR used a new operation below Zeration for defining ln(△a) and created new units for other operations, and he mentioned that there is an infinite amount of new sets of numbers. It would be interesting to see if this can be generalized in some way.


I still have unanswered questions about this algebraic structure:

- How can we extend Zeration to the complex set? What would (3*i) ° 4 be equal to? What about j ° 2 ?
- What is the proof for Zeration's commutativity? I saw somewhere on the forum that there was a proof for that but I couldn't find it.
- What about other hypothetical extensions I didn't mention?
- Have I made any mistakes? What did other people discover?


Rayanso]]>
On this forum I discovered GFR's post and I looked at the paper, and I also used the wayback machine to download an old paper from 1996 on hyper-operations from KAR's website.
If you are not familiar with Zeration and Deltation I suggest that you read the file that GFR attached in his post.

One question I was asking myself was what 0.5 * (△0) would be equal to, and if it could be reduced to a single number. In KAR's paper, this is resolved by introducing a new unit "j", like the imaginary unit for (-1)^(0.5). We then have 2*j =  △0. The operation 0.5 * (△0) also has 2 solutions: △j and j.

We still don't know what 0.35801 * (△0), or with any other arbitrary coefficient, would equal to. I made a formula for any real number as a coefficient for what this would be equal to.

I came up with x * (△0) = j*2*(2n+1)*x with n being an integer. I verified this formula with (-1)^a = b^(△0 * a)

Verification: for (-1)^(1/2) we have b^(△0 * 1/2) = {b^j ; b^(△j)}, so we must have b^j = i
b is a real number that is not zero.

For (-1)^(1/3) we know that there will be 3 solutions, therefore, △0 * 1/3 will have 3 solutions. These are: △0 ; 2/3 * j ; -2/3 * j. All other solutions can be reduced to one of these 3.

We also have an interesting rule that completes one that already exists:

a^0 = 1
a^(△0) = -1
a^j = i
a^(△j) = -i

This makes sense because the absolute value is 1 in every case, and the exponent is linked in some way to the number 0. 
If we have b = a*(△0), for any real a. c^b will always have an absolute value of 1. 

I also discovered the rule △j = -j, which I think KAR has not mentioned in his paper.

Proof:

△j + △j = 2*△j = △0
△j + (-j) = △0 + j - j = △0
(-j) + (-j) = -(2*j) = -△0 = △0

This is as far as I went and there are many gaps in the algebraic structure, KAR used a new operation below Zeration for defining ln(△a) and created new units for other operations, and he mentioned that there is an infinite amount of new sets of numbers. It would be interesting to see if this can be generalized in some way.


I still have unanswered questions about this algebraic structure:

- How can we extend Zeration to the complex set? What would (3*i) ° 4 be equal to? What about j ° 2 ?
- What is the proof for Zeration's commutativity? I saw somewhere on the forum that there was a proof for that but I couldn't find it.
- What about other hypothetical extensions I didn't mention?
- Have I made any mistakes? What did other people discover?


Rayanso]]> 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.
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.
i hope someone finds a faster way to implement tetration, pentation, etc. in desmos.]]> https://tetrationforum.org/showthread.php?tid=1805 Sat, 30 Aug 2025 14:08:42 +0000 Alex Zuma 2025]]> https://tetrationforum.org/showthread.php?tid=1805 my birthday: january 21, 2008
gender: male
pronouns: he/him/his
i was born in romania and i live in romania.
i like video editing, gameplays, advanced mathematics, and googology.
i also like to create graphs on desmos.
a few graphs i made on desmos:
base e hyperoperations: https://www.desmos.com/calculator/ueqda2tbk8
base 2 hyperoperations : https://www.desmos.com/calculator/q3gbpkftz8
base 10 hyperoperations : https://www.desmos.com/calculator/vz8uwgwgql
base 3 hyperoperations : https://www.desmos.com/calculator/l4i9w4b2am
base 4 hyperoperations : https://www.desmos.com/calculator/mevebbamf9
base 1.7 hyperoperations : https://www.desmos.com/calculator/dl1usyacyw
base 1.5 hyperoperations : https://www.desmos.com/calculator/wluijn05la
the cheta function: https://www.desmos.com/calculator/ypj0mvflmd
the xth super root of 2: https://www.desmos.com/calculator/akhnu1rpxo
the xth super root of x and infinite pentation: https://www.desmos.com/calculator/heiz8lyy3u
half tetrations: https://www.desmos.com/calculator/aettjlb2qp
the tetration of the square root of 2: https://www.desmos.com/calculator/qyhpxmyieu
tetration base eta: https://www.desmos.com/calculator/8obpqlvmwo
half-exponential function approximated by a polynomial: https://www.desmos.com/calculator/kb813o3cdx
the tetration of pi: https://www.desmos.com/calculator/udxupmmygc
i hope you enjoy seeing these graphs as much as i enjoyed making them!]]> my birthday: january 21, 2008
gender: male
pronouns: he/him/his
i was born in romania and i live in romania.
i like video editing, gameplays, advanced mathematics, and googology.
i also like to create graphs on desmos.
a few graphs i made on desmos:
base e hyperoperations: https://www.desmos.com/calculator/ueqda2tbk8
base 2 hyperoperations : https://www.desmos.com/calculator/q3gbpkftz8
base 10 hyperoperations : https://www.desmos.com/calculator/vz8uwgwgql
base 3 hyperoperations : https://www.desmos.com/calculator/l4i9w4b2am
base 4 hyperoperations : https://www.desmos.com/calculator/mevebbamf9
base 1.7 hyperoperations : https://www.desmos.com/calculator/dl1usyacyw
base 1.5 hyperoperations : https://www.desmos.com/calculator/wluijn05la
the cheta function: https://www.desmos.com/calculator/ypj0mvflmd
the xth super root of 2: https://www.desmos.com/calculator/akhnu1rpxo
the xth super root of x and infinite pentation: https://www.desmos.com/calculator/heiz8lyy3u
half tetrations: https://www.desmos.com/calculator/aettjlb2qp
the tetration of the square root of 2: https://www.desmos.com/calculator/qyhpxmyieu
tetration base eta: https://www.desmos.com/calculator/8obpqlvmwo
half-exponential function approximated by a polynomial: https://www.desmos.com/calculator/kb813o3cdx
the tetration of pi: https://www.desmos.com/calculator/udxupmmygc
i hope you enjoy seeing these graphs as much as i enjoyed making them!]]>