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 Generalized arithmetic operator hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 03/11/2014, 03:52 AM (This post was last modified: 03/11/2014, 03:54 AM by hixidom.) Hello forum. So I've finally hit a dead end in my hyperoperator theory. "Dead end" as in I fear I've discovered that there is, in general, no closed for for [x]a, where [x] is my hyperoperator (with argument x) and a is the argument. So this is what I've got so far: Amateur document (The real meat of the theory starts in Section 4) Crude summary Define a binary operation, $a[x]b$, where $a[1]b=a+b$, $a[2]b=a\cdot b$, $a[3]b=a^b$ and so on. My goal is to get to a point where I can evaluate $a[x]b$ for $x\in C$. Next, define a unary operation, $[x]a$. Denote iteration of the binary and unary operators via a superscript: e.g. $a[x]^3 b=((a[x]b)[x]b)[x]b$ $[x]^3 a=[x][x][x]a$ Next, establish the following axioms: A0: $a[x]^0 b=a$ A1: $[x]a\equiv a[x]a$ A2: $[x]^n [x]^m a=[x]^{n+m}a$ (perhaps this is not really an axiom based on the way I have defined iteration of the unary operator) A3: $[x]^n a=a[x]2^n$ From these "axioms" the following relations, among others, can be proven (see the document for derivations): $a[4]b$ is not tetration (sorry forum) $(a[x]b)[x]c=a[x](b\cdot c)$ $a[x]^n b=a[x]b^n$ $(a[x]b)[x]^{-1} b=a$ (definition of inverse) $[x]2=[x+1]2$ (2 is a fixed point, explaining why $2+2=2\cdot 2=2^2$, etc.) The problem: I find that $[x+y]a=f_y^{log_2 a}(a)$, where $f_y(a)$ is the yth super-iteration of $f_1(a)\equiv [x]a$. What do I mean by "super-iteration" and how do I come to that conclusion? It's in Section 4.3 of the document, but here it is again anyway: Given the axioms, we can find that $[x]^{log_2 a}a=[x+1]a$. Define $f_1(a)\equiv [x]a$. Now we can iterate both sides $log_2 a$ times to obtain $f_1^{log_2 a}(a)=[x]^{log_2 a}a=[x+1]a$. Now we must define $f_2(a)\equiv f_1^{log_2 a}(a)$, and repeat the $log_2 a$ iteration to obtain $f_2^{log_2 a}(a)=[x+1]^{log_2 a}a=[x+2]a$. Repeating this algorithm y times results in $f_y^{log_2 a}(a)=[x+y]a$. I know that the "super-iteration" method is necessary because I've tried other more naive derivations of $[x+y]a$ which turned out to be wrong when testing various combinations of x and y (e.g. $[x+y]a$ should be the same for (x,y)=(0,1) and (x,y)=(1,0)). So anyways, I'm putting this out here on the forum because super-iteration seems to be an unmanageable concept to me, and I have not been able to find a way to avoid it. I hope that someone with more experience in this area of mathematics will have some idea of how to manipulate the super-iteration into something simpler or how to relate $[x]a$ and $[x+y]a$ without the need for super-iteration. Ideally, the end result is a closed form expression for $[x]a$, where x can be any complex number. I'm using the term "super-iteration" because it is used on another thread on the forum (which, btw, is the only place on the internet where I could find information on anything similar to what I'm encountering). Thanks for reading. I know that this is my problem, but if anyone finds it interesting enough to think about, I would appreciate any comments on the validity of what I have so far and suggestions on how to proceed. Thanks again. hixidom « Next Oldest | Next Newest »

 Messages In This Thread Generalized arithmetic operator - by hixidom - 03/11/2014, 03:52 AM RE: Generalized arithmetic operator - by JmsNxn - 03/11/2014, 03:15 PM RE: Generalized arithmetic operator - by hixidom - 03/11/2014, 06:24 PM RE: Generalized arithmetic operator - by MphLee - 03/11/2014, 10:49 PM RE: Generalized arithmetic operator - by hixidom - 03/11/2014, 11:20 PM RE: Generalized arithmetic operator - by MphLee - 03/12/2014, 11:18 AM RE: Generalized arithmetic operator - by JmsNxn - 03/12/2014, 02:59 AM RE: Generalized arithmetic operator - by hixidom - 03/12/2014, 04:37 AM RE: Generalized arithmetic operator - by MphLee - 03/12/2014, 06:19 PM RE: Generalized arithmetic operator - by hixidom - 03/12/2014, 06:43 PM RE: Generalized arithmetic operator - by tommy1729 - 03/21/2014, 10:31 PM RE: Generalized arithmetic operator - by hixidom - 03/22/2014, 12:06 AM RE: Generalized arithmetic operator - by tommy1729 - 03/22/2014, 12:13 AM RE: Generalized arithmetic operator - by hixidom - 03/22/2014, 12:42 AM RE: Generalized arithmetic operator - by hixidom - 06/11/2014, 05:10 PM

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