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Generalized arithmetic operator
#1
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, , where , , and so on. My goal is to get to a point where I can evaluate for . Next, define a unary operation, . Denote iteration of the binary and unary operators via a superscript:
e.g.


Next, establish the following axioms:
A0:

A1:

A2: (perhaps this is not really an axiom based on the way I have defined iteration of the unary operator)

A3:

From these "axioms" the following relations, among others, can be proven (see the document for derivations):

is not tetration (sorry forum)





(definition of inverse)

(2 is a fixed point, explaining why , etc.)

The problem:
I find that , where is the yth super-iteration of .

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 . Define . Now we can iterate both sides times to obtain

.

Now we must define , and repeat the iteration to obtain

.

Repeating this algorithm y times results in

.

I know that the "super-iteration" method is necessary because I've tried other more naive derivations of which turned out to be wrong when testing various combinations of x and y (e.g. 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 and without the need for super-iteration. Ideally, the end result is a closed form expression for , 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
Reply


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 hixidom - 03/22/2014, 12:06 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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