Zeration

[bo198214]

This multiple way (quadruple, not ... octuple) is of the inductive and not of the deductive type. I hope that everybody would be patient enough to read it, without ... fainting,

At least I do it!

deciding to go to the Foreign Legion or (BO) organizing a metaphorical Srafspedition for the democratic elimination of ... somebody from this Forum.

We dont eliminate anyone from the forum. And regarding opinions its not so much a democratic principle, but rather the mathematical principle similar to Ockham's razor: Unnecessary assumptions are eliminated.

Unless (there is always an ... unless) the Mother Law is not alone.
...
we should be prepared to find other additional conditions, accompanying and supporting the Mother Law.

You can not supplement the mother law, it alone prescribes that a[0]b=b+1 for all b.

Pillar 2 - The Ackermann Pillar.

The Ackermann function as you describe it is a modification of the original Ackermann function with 3 arguments
[Ackermann, Zum Hilbertschen Aufbau der reellen Zahlen, Math. Ann. 99 (192, 118-133]
which is exactly our hierarchy a[n]b except that it starts with the 0th operation being the addition instead of the first one.

The modification was used to easier prove statements like the existence of recursive but not primitive recursive functions. And no wonder does it lead also here to the conclusion that a[0]b=b+1.

Pillar 3 - The Hyper-roots.
...
y = x /[s]2 ---> y <= (y[s-1](y[s]\ x)) /[s-1]2.

I think it should read (y[s-2](y[s-1]\ x)) /[s-1]2 right?

So let us see why this formula actually computes the value of x/[s]2.
It is in iteration formula, so if it has a limit y then
y = (y[s-2](y[s-1]\ x)) /[s-1]2

If we further assume that all arguments are in its bijectivity domain this formula is equivalen to:

y [s-1] 2 = y [s-2] ( y [s-1]\ x )
y [s-2] (y[s-1]1) = y [s-2] (y [s-1]\ x )
y[s-1]1=y [s-1]\ x
y [s-1] (y[s-1] 1) = x

And now under the assumption that y[s]0=1:

y [s] 2 = y [s] (1+1+0) = y[s-1](y[s-1](y[s]0)) = x
y = x [s]/ 2

However the assumption is wrong for s=2, y[2]0=0.

Konstantin Rubtsov (Rubcov) knows a complicated, but very "clean" demonstration of the commutativity of zeration,

However he did not cleanly state from what laws this commutativity follows. Surely not from the mother law.

For the moment, I keep the Faith, thinking that, after all, we could consider that statement as part of a postulated axiom (BO ipse, in a moment of ... weakness, dixit!).

Of course if you could derive your solution by the assumption of commutativity that would be fine, however it still contradicts the mother law and I think not even commutativity would make it unique.

Pillar 4 - The Hyper-means.

The hyper means (a[n]a)/[n+1] 2 = a
follow directly from the assumption that a[n]a=a[n+1]2 which is equivalent to a[n+1]1=a.

a ^ a = a # 2
a * a = a ^ 2
a + a = a * 2
........
as well as, more particularly (for a = 2), the almost "holy" tetragonal equality:
.... 2 + 2 = 2 * 2 = 2 ^ 2 = 2 # 2 = ..... 2 [s] 2 .... = 4 !!!!!!

Yes, but Gianfranco ( ), all those laws follow from the mother law:
2[s+1]2 = 2[s](2[s+1]1) = 2[s]2 = 4, only for 2[s+1]1=2!
Not for s=0, as 2[1]1=3\( \neq \)2.

Asserting that a[0]a=a+2 is a bit like asserting that a[1]1=a.
Do you see the similarity?
Assume we had the hierarchy beginning at multiplication [2] and we try to find the operation below [2], i.e. the addition [1].
Then you see, oh, a[n]1=a for all n>1, hence you would be tempted to propose that a[1]1=a though this contradicts the mother law:
1\( \neq \)2=1[2]2=1[1](1[2]1)=1[1]1

In the same way you see, oh, a[n]a=a[n+1]2 for all n>0 and you are tempted to propose that a[0]a=a+2 though this contradicts the mother law (thanks for giving us this term!).

I have some doubts concerning its unicity, but this happens in the best families, such as the Euler's Gamma function as extension of the factorial.

Absolutely not, the Euler Gamma function is unique under the condition of logarithmic convexity.
However I never saw any condition that would make KAR's zeration unique. I even showed that it is not unique under certain strong conditions and provided a counter example (different from a[0]b=b+1).

The Neutral Elements - What somebody calls the left/right unit elements. Let us consider the following general functional equation, including the definition of a fixpoint of the type x = f(x):
x = a[s]x <---> x = a[s+1]oo, which means:
a = x /[s]x <---> a = x /[s+1]oo (inverses, of the root types)

The implementation of that for ranks 3, 2, 1, 0 gives:
x = a^x <---> x = a#oo
a = x-rt x <---> a = oo-srt x = x^(1/x) (no neutral element for rank 3)

x = a*x <---> x = a^oo
a = x/x = 1 <---> a = oo-rt x = x^0 = 1

x = a+x <---> x = a*oo
a = x-x = 0 <---> a = x/oo = 0

x = a°x <---> x = a+oo
a = xçx <---> a = x-oo = -oo (ç stands here for the delta symbol, inverse of "°")

But even the neutral elements dont fall from sky, they can also derived from the mother law, though then you see the real law:

If \( \lim_{t\to\infty} \) x /[n+1] t = a then a[n]x=x[n+1]1 (or equivalently (x[n+1]1) /[n] x = a).

And a=x/[n]x only for n>0, because there x[n+1]1=x! You see, even though I didnt intended it, that this is exactly the errornous conclusion/induction to go from x[m]1=x for all m>1 to x[1]1=x, if you assert that also a[0]x=x instead of a[0]x=x[1]1=x+1 (setting n=0 in the above law).

Derivation of the above law:
a[n]x =\( \lim_{t\to\infty} \)(x /[n+1] t) [n] x

(x /[n+1] t) [n] x = (x /[n+1] t) [n] ((x /[n+1] t)[n+1]t) = (x /[n+1] t) [n+1] (t+1) = x [n+1] 1

then a[n]x=\( \lim_{t\to\infty} \) (x[n+1]1)=x[n+1]1.

Conclusion: each mentioned pillar (without the incorrectly made generalization to zeration) is a (mathematically strict) consequence of the mother law. The general form of the pilars are:

Pillar 3 - The Hyper-roots.
If \( y \) is the limit of the sequence \( y_{n+1}=(y_n[s-2](y_n[s-1]\backslash x)) /[s-1]2 \) then \( y[s-1](y[s-1]1)=x \).
If (and only if) \( y[s]0=1 \) then \( y=x/[s]2 \).

So the conclusion that \( y_{n+1}=(y_n[0](y_n [1]\backslash x)) /[1] 2 \) tends to \( x/[2]2 \) is wrong, as \( y[2]0=0 \).

Pillar 4 - The Hyper-means.
The general law is a[s]2=a[s-1](a[s]1).
If (and only if) a[s]1=a then a = (a[s-1]a)/[s] 2.

this is not satisfied for s=1, a[1]1=a+1\( \neq \)a, and hence the conclusion a=(a[0]a)/[1] 2 is wrong.

Pillar 5 - The hyper root limits
If \( \lim_{t\to\infty} \) x /[n+1] t = a then a[n]x=x[n+1]1 (or equivalently (x[n+1]1) /[n] x = a).
If (and only if) x[n+1]1=x then a[n]x=x.

x[1]1\( \neq \)x hence the conclusion that a[0]x=x is wrong.

In turn all the by you mentioned pillars completely *support* the definition a[0]b=b+1.

Gianfranco, I know you and KAR invested a lot into the development of your zeration. But sometimes its just time to let go.