06/10/2010, 06:42 AM
Actually there were some heated debates on this forum about zeration.
I dont recollect all the posts but this thread or ths thread may give a good impression.
I fully agree with your derivation that all operations \( \le 0 \) are just the increment operation. It follows from the general rule you give to build the hyperoperations ladder.
Actually not much result results were collected about operations with fractional ranks or even complex ranks. Your question appeared already in this thread. My idea about it is (I posted it once but couldnt find it anymore) to express the hyperoperations as repetition of an operator and then take fractional powers of that operator. There is a theory about fractional powers of operators, see here.
For example if we only care about the superfunctions x -> b [n] x, then as you mentioned we have the equation
b [n+1] (x+1) = b [n] (b [n+1] x)
that means to get the n+1-th superfunction g(x) = b [n+1] x from f(x) = b [n] x
we create some operator S (successor operator) that satisfies
S[f](x+1)=f(S[f](x)).
Now this operator S may be expressed as an infinite matrix transforming the powerseries coefficients of f, or whatever, and we can take powers of it:
\( f_0(x)=b+x \)
\( S[f_0](x) = bx \)
\( S^2[f_0](x) = b^x \)
...
and then we can also take perhaps fractional or complex powers \( S^z \).
I dont recollect all the posts but this thread or ths thread may give a good impression.
I fully agree with your derivation that all operations \( \le 0 \) are just the increment operation. It follows from the general rule you give to build the hyperoperations ladder.
Actually not much result results were collected about operations with fractional ranks or even complex ranks. Your question appeared already in this thread. My idea about it is (I posted it once but couldnt find it anymore) to express the hyperoperations as repetition of an operator and then take fractional powers of that operator. There is a theory about fractional powers of operators, see here.
For example if we only care about the superfunctions x -> b [n] x, then as you mentioned we have the equation
b [n+1] (x+1) = b [n] (b [n+1] x)
that means to get the n+1-th superfunction g(x) = b [n+1] x from f(x) = b [n] x
we create some operator S (successor operator) that satisfies
S[f](x+1)=f(S[f](x)).
Now this operator S may be expressed as an infinite matrix transforming the powerseries coefficients of f, or whatever, and we can take powers of it:
\( f_0(x)=b+x \)
\( S[f_0](x) = bx \)
\( S^2[f_0](x) = b^x \)
...
and then we can also take perhaps fractional or complex powers \( S^z \).