Zeration
#91
Quote:Then, why not make a zerated to the a equal a plus 3 halves as a compromise?

Because that doensn't make any sense.
I know that to someone who is not familiar with the material, all the disccussion about zeration sounds meaningless and arbitrary, but it isn't. I encourage you to read the literature about that. Rubtsov and Romerio gave great motivation and heuristics in their papers. There is also Cesco Reale's article, and many rich threads on this forum. Look for them. The keyword here is mother law.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#92
(06/30/2022, 11:32 PM)MphLee Wrote:
Quote:Then, why not make a zerated to the a equal a plus 3 halves as a compromise?

Because that doensn't make any sense.
I know that to someone who is not familiar with the material, all the disccussion about zeration sounds meaningless and arbitrary, but it isn't. I encourage you to read the literature about that. Rubtsov and Romerio gave great motivation and heuristics in their papers. There is also Cesco Reale's article, and many rich threads on this forum. Look for them. The keyword here is mother law.
The Mother Law would make a°a=a+1. The Grand-Mother Law would make a°a=a+2. Why not make a°a=a+1.5, As a compromise zeration?
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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#93
Again, because that doesn't make any sense. If you do that you are imposing the failure of both the laws, something that, per se, is a folish move if you don't come up with a better law. And the purpose of the laws is to add constraints to the functional equations in order to fully determine solutions to of hyperoperations where the zeroth-rank is a total function. Again, you need to read more literature and put some effort.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#94
I want to point to this other thread where a novel possible definition for operation preceding addition is proposed by user Danteman163.
It seems to me that this idea was never been described before.

https://math.eretrandre.org/tetrationfor...32#pid9332

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#95
(12/28/2022, 01:28 PM)MphLee Wrote: I want to point to this other thread where a novel possible definition for operation preceding addition is proposed by user Danteman163.
It seems to me that this idea was never been described before.

https://math.eretrandre.org/tetrationfor...32#pid9332

Hmm I actually really like that. Especially his description. Is that a viable operator though...? Personally I still like the \(\Gamma\) function interpretation.

\[
a \langle 0 \rangle b = a + b\\
\]

Then:

\[
\begin{align}
a \langle -1 \rangle b &= a +1\\
a \langle -2 \rangle b &= a+1\\
a \langle -3 \rangle b &= a+1\\
&\vdots
\end{align}
\]

But, there's singular behaviour at each of these points. And that small perturbations in the hyper operational indexes are wild and destructive. Additionally, the action of going back becomes idempotent. There isn't a 1-1 in the mapping \(\langle q \rangle \to \langle q-1\rangle\). But this only happens at the singular points at \(-1,-2,-3,...\). You need local data near \(q \approx -2\) to pull back to addition again. Despite \(q =-2\) alone making \(q=-1\) addition when you try to naively map forward \(\langle q \rangle \to \langle q+1\rangle\). Where doing this naively just iterates the predecessor. Doing it more difficultly requires iterating a neighborhood of the predecessor.

I apologize. I like this idea from zeration. But honestly. It should be successorship all the way down, but at each singularity, there is different behaviour in a neighborhood of each successorship.
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#96
I agree, in fact I'm working on Goodstein-like solutions on my quest for naturality.
But there can be a non-trivial zerations zoo of solutions. And this thread is all about that alternative exotic landscape. One day we will be able to appreciate better its taste and meaning I believe, eg. its relationship to tropical math.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#97
(01/05/2023, 08:53 PM)MphLee Wrote: I agree, in fact I'm working on Goodstein-like solutions on my quest for naturality.
But there can be a non-trivial zerations zoo of solutions. And this thread is all about that alternative exotic landscape. One day we will be able to appreciate better its taste and meaning I believe, eg. its relationship to tropical math.

Oh yes! For sure Mphlee. There's a whole zeration zoo. Same way there is a zoo of "interpolations of the factorial". But they're all garbage other than \(\Gamma\), lmao Big Grin .
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#98
(01/05/2023, 08:53 PM)MphLee Wrote: I agree, in fact I'm working on Goodstein-like solutions on my quest for naturality.
But there can be a non-trivial zerations zoo of solutions. And this thread is all about that alternative exotic landscape. One day we will be able to appreciate better its taste and meaning I believe, eg. its relationship to tropical math.

Hello MphLee!

I'm interested in negative-rank hyperoperations, and I posted this on the googology fandom of japan. I tried to solve the equation for bi-infinite sequence \((\star_r: \mathbb{Z}_{\geq 2} \times \mathbb{Z}_{\geq 2} \to \mathbb{Z}_{\geq 2})_{r = -\infty}^{\infty}\) as follows:
\[
    \forall m \forall r \forall n; m \star_{r+1} (n+1) = m \star_r (m \star_{r+1} n) \land m \star_{r+1} 2 = m \star_r m \land m \star_1 n = m+n.
\]
I proofed the existence of solutions and found one, \(\star_0 = \heartsuit\) such that
\[
    m \heartsuit n =
    \begin{cases}
        n & \text{if}\; 2 \leq n \leq m-1 \\
        n+2 & \text{if}\; n = m \\
        n & \text{if}\; n = m+1 \\
        n+1 & \text{if}\; n \geq m+2.
    \end{cases}
\]
Then all negative elements are uniquely determined as
\[
    m \star_{-1} n =
    \left\{
    \begin{array}{lll}
        & & \text{if \(m = 2\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        n+1 & = 2 & n = 1 \\
        n+2 & = 4 & n = 2 \\
        n+2 & = 5 & n = 3 \\
        n-1 & = 3 & n = 4 \\
        n+1 & \geq 6 & n \geq 5 \\
        & & \text{if \(m = 3\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        n+2 & = 3 & n = 1 \\
        n+3 & = 5 & n = 2 \\
        n-1 & = 2 & n = 3 \\
        n+2 & = 6 & n = 4 \\
        n-1 & = 4 & n = 5 \\
        n+1 & \geq 7 & n \geq 6 \\
        & & \text{if \(m \geq 4\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        m & = m & n = 1 \\
        n+1 & \geq 3, \leq m-1 & 2 \leq n \leq m-2 \\
        n+3 & = m+2 & n = m-1 \\
        2 & = 2 & n = m \\
        n+2 & = m+3 & n = m+1 \\
        n-1 & = m+1 & n = m+2 \\
        n+1 & \geq m+4 & n \geq m+3
    \end{array}
    \right.
\]
\[
    m \star_{-2} n =
    \left\{
    \begin{array}{lll}
        & & \text{if \(m = 2\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        n+1 & = 2 & n = 1 \\
        n+2 & = 4 & n = 2 \\
        n+3 & = 6 & n = 3 \\
        n+1 & = 5 & n = 4 \\
        n-2 & = 3 & n = 5 \\
        n+1 & \geq 7 & n \geq 6 \\
        & & \text{if \(m = 3\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        n+2 & = 3 & n = 1 \\
        n+4 & = 6 & n = 2 \\
        n+2 & = 5 & n = 3 \\
        n+3 & = 7 & n = 4 \\
        n-3 & = 2 & n = 5 \\
        n-2 & = 4 & n = 6 \\
        n+1 & \geq 8 & n \geq 7 \\
        & & \text{if \(m = 4\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        n+3 & = 4 & n = 1 \\
        n+5 & = 7 & n = 2 \\
        n+3 & = 6 & n = 3 \\
        n-1 & = 3 & n = 4 \\
        n+3 & = 8 & n = 5 \\
        n-4 & = 2 & n = 6 \\
        n-2 & = 5 & n = 7 \\
        n+1 & \geq 9 & n \geq 8 \\
        & & \text{if \(m \geq 5\) and} \\
        n+1 & \leq 1 & n \leq 0 \\
        m & = m & n = 1 \\
        m+3 & = m+3 & n = 2 \\
        n+1 & \geq 4, \leq m-1 & 3 \leq n \leq m-2 \\
        n+3 & = m+2 & n = m-1 \\
        3 & = 3 & n = m \\
        n+3 & = m+4 & n = m+1 \\
        2 & = 2 & n = m+2 \\
        n-2 & = m+1 & n = m+3 \\
        n+1 & \geq m+5 & n \geq m+4
    \end{array}
    \right.
\]
...
If the domain and codomain are limited to \(\mathbb{Z}_{\geq 2}\), we can interpolate the hyperoperation sequence.


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