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
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