# Tetration Forum

Full Version: Zeration
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4 5 6 7 8 9
(03/23/2015, 01:39 PM)marraco Wrote: [ -> ]
(03/21/2015, 11:11 PM)tommy1729 Wrote: [ -> ]a*b = b*a

That would mean
b*(a-1)=a*(b-1)

but if "-" is an inverse operator of [q], such that x-y is defined as x-y=x[q]-y, and -y is defined as y[q]-y=N(q,y) then

for [q]=product, "-" would be division, so
Ure misunderstanding the expression $f^{n}(x)$

In fact Tommy tryes to replace g and f with "add_b" and "add_1" and the interpretation becomes the following

Quote:General case $f^{a - N_q}(N_q[q]b) = g^{b - N_q}(N_q[q]a)$

- Case q=1 with $[1]=+$, $f(x)=g(x)={\rm add}_1(x)=x+1$ and $N_1=0$

${\rm add}_1^{a - N_1}(N_1[1]b) = {\rm add}_1^{b - N_1}(N_1[1]a)$
${\rm add}_1^{a - 0}(0+b) = {\rm add}_1^{b -0}(0+a)$
${\rm add}_1^{a }(b)=a+b =b+a= {\rm add}_1^{b }(a)$

- Case q=2 with $[2]=\cdot$, $f(x)={\rm add}_b(x)=b+x$, $g(x)={\rm add}_a(x)=a+x$ and $N_2=1$

${\rm add}_b^{a - N_2}(N_1[2]b) = {\rm add}_a^{b - N_2}(N_2[2]a)$
${\rm add}_b^{a - 1}(1 \cdot b) = {\rm add}_1^{b -1}(1\cdot a)$
$(b+...+b)_{a-1 times}+(b)=(a+...+a)_{b-1 times}+(a)$
$b(a-1)+(b)=a(b-1)+(a)$
$(ba-b)+(b)=ba=ab=(ab-a)+(a)$

- Case q=0 with $[0]=\circ$ and $N_0=-\infty$, but in this case how we get the following

$f^{a -(-\infty)}((-\infty)[0]b) = g^{b - (-\infty)}((-\infty)[0]a)$

$f^{+\infty}(b) = g^{+\infty}(a)$

Quote:-∞ is used on all the basic operations:

the neutral of addition is ÷∞=1/∞=0
the neutral of product is ∞√=∞√n=n^÷∞=1
the neutral of exponentiation is $^{-\infty}n=n^{\frac{1}{n}}$

from product viewpoint, all numbers smaller than 0 are transfinite.
from exponentiation of n viewpoint, all numbers smaller than 1 are transfinite.
from tetration of n viewpoint, all numbers smaller than $n^{\frac{1}{n}}$ are transfinite.
Agree with this last point...
It really deserves some attention imho.
(03/21/2015, 11:11 PM)tommy1729 Wrote: [ -> ]f and g are some function :

f^[a - N_q](N_q[q]b) = g^[b - N_q](N_q[q]a)

(03/23/2015, 02:31 PM)MphLee Wrote: [ -> ]
Quote:General case $f^{a - N_q}(N_q[q]b) = g^{b - N_q}(N_q[q]a)$
maybe this one is clearer?
$f^{a}(N_{(q,a)}) = g^{b}(N_{(q,b)})$
$f(x)=x {\small{[q-1]}} b$
$g(x)=x {\small{[q-1]}} a$

I added a subindex to the neutral, because many operations have a different neutral for each number.

For example, right bracket tetration:
$(..((a_1)^{a_2})^{a_3}^ \,...)^{a_b}=a^{a^{b-1}}$, where
$f_{(x)}=x {\small{[3]}} b = x^b$
$g_{(x)}=x {\small{[3]}} a = x^a$

...looks non intuitive. Unnecessarily complicated.
But each "a" has a neutral $N_{()4,a}\,=\, a^{a^{-1}}$, which simplifies the notation:

$(..((({N_a)}^{a_1})^{a_2})^{a_3}^ \,...)^{a_b}={N_a\,}^{a^{a^b}}$

(03/21/2015, 11:11 PM)tommy1729 Wrote: [ -> ]f^[a - oo](...) does not make sense as a nonconstant function.
a+b = b+a

Hello!

It is interesting that I am also from Russia just as like K.A. Rubcov, and since 2004 me and one my friend, we have often been thinking and talking about the zeration. We seriosely tried to resolve this problem rationally. Our discussion on the matter sometimes stopped, or sometimes continued along all these years. This friend had even found on the web an article of K.A. Rubcov in Russain language, and we made some notes on it.

But we use another symbol for the operation: "@", i. e. a@a = a+2; and we call it, in Russian, with a word "набирание" which is close to English words "making a set", "making a collection". We have chosen this word because of it's psycological sense: before additing some numbers to each other, we must imagine them together, in a collection, in a set.

We two tried to satisfy all the restrictions presented here:

1) a@a = a+2 like a+a = a*2
2) a+(b@c) = a+b@a+c like a*(b+c) = a*b+a*c
3) a@b = b@a just we want it be so
4) a@b < a+b < a*b for all a not equal b (probably)

And I think that we achieved a good solution. Now I am preparing an article about it and thinking of where to publish it.

If you are still interested in it or just do not deny the possibility of a good solution, then I could hurry up with an article.
(05/28/2015, 11:12 PM)Stanislav Wrote: [ -> ]If you are still interested in it or just do not deny the possibility of a good solution, then I could hurry up with an article.

Of course we are interested. We are the most interested people on Internet.
(05/29/2015, 01:33 AM)marraco Wrote: [ -> ]Of course we are interested. We are the most interested people on Internet.

OK. I've already written an article with my ideas and made a program for calculations with zeration. The article needs to be translated into English and to be converted into .pdf from .doc, and the program is to be made as .dll or .exe from .xls.
Hi Stanislav!

Since the 2014 i've been working with G.F. Romerio, a Rubtsov's strict collaborator. I'm studying Rubtsov's and Romerio's publications since 2006, when their first document where published. They made alot of stuff under the name "K.A.Rubtsov and G.F. Romerio" and updated all their activities on the Wolfram Research's forum called "The NKS Forum" (see: http://forum.wolframscience.com/forumdis...&forumid=7)

Are you aware of their new developements on Zeration?

Also here on the Tetration forum there are interesting threads:
http://math.eretrandre.org/tetrationforu...ostid=6819
I find this one really advanced on the state of this topic (Zeration).
(06/03/2015, 01:40 PM)MphLee Wrote: [ -> ]Hi Stanislav!

Since the 2014 i've been working with G.F. Romerio, a Rubtsov's strict collaborator. I'm studying Rubtsov's and Romerio's publications since 2006, when their first document where published. They made alot of stuff under the name "K.A.Rubtsov and G.F. Romerio" and updated all their activities on the Wolfram Research's forum called "The NKS Forum" (see: http://forum.wolframscience.com/forumdis...&forumid=7)

Are you aware of their new developements on Zeration?

Also here on the Tetration forum there are interesting threads:
http://math.eretrandre.org/tetrationforu...ostid=6819
I find this one really advanced on the state of this topic (Zeration).

MphLee,

I've read some two articles af G.F. Romerio & K.A. Rubcov, one of which is of 2006. In some aspects they achieve similar basic results, such as idempotent -infinity. But generally I found a different way to issue this matter. Anyway, now I am translating my paper into English.
(06/03/2015, 01:40 PM)MphLee Wrote: [ -> ]If you are not I suggest to read this thread

PS: Also there are not translation of Reale's paper yet but it is about the binary operation $a\odot b={\max}(a,b)+1$ and a new set of numbers that is defined by the pairs of real numbers up to a special equivalence relation.