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 Zeration MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 03/23/2015, 02:31 PM (03/23/2015, 01:39 PM)marraco Wrote: (03/21/2015, 11:11 PM)tommy1729 Wrote: a*b = b*a "add+b"^[a-1](1*b) = "add+a"^[b-1](1*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, soUre 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. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/24/2015, 11:29 AM (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 "add+1"^[a](0) = "add+1"^[b ](0) Stanislav Junior Fellow Posts: 6 Threads: 1 Joined: May 2015 05/28/2015, 11:12 PM (This post was last modified: 05/28/2015, 11:15 PM by Stanislav.) 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. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 05/29/2015, 01:33 AM (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. I have the result, but I do not yet know how to get it. Stanislav Junior Fellow Posts: 6 Threads: 1 Joined: May 2015 05/29/2015, 09:06 PM (This post was last modified: 05/29/2015, 09:07 PM by Stanislav.) (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. MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 06/03/2015, 01:40 PM 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? If you are not I suggest to read this thread http://forum.wolframscience.com/showthre...eadid=1984 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). MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ Stanislav Junior Fellow Posts: 6 Threads: 1 Joined: May 2015 06/04/2015, 06:44 AM (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? If you are not I suggest to read this thread http://forum.wolframscience.com/showthre...eadid=1984 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. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 06/04/2015, 08:44 PM (This post was last modified: 06/04/2015, 10:48 PM by marraco.) (06/03/2015, 01:40 PM)MphLee Wrote: If you are not I suggest to read this thread http://forum.wolframscience.com/showthre...eadid=1984I have great difficulty understanding the papers in Italian. In particular about the "Escherian numbers". I have the result, but I do not yet know how to get it. MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 06/05/2015, 09:10 PM There are not many papers in italian language, C. Reale's one on the Escherian number is the only text in italian, the others are all in english. 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. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ Stanislav Junior Fellow Posts: 6 Threads: 1 Joined: May 2015 09/09/2015, 10:04 PM Dear Colleagues, unfortunately, I have overestimated my free time and haven't not yet finished to write a paper about my considerations for zeration, my conclusions for calculating it. But I do my best to hurry up with it. There stays a gap for applying my method to complex numbers, and yet I don't know how to deal with it. « Next Oldest | Next Newest »

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