Personally, I've always used the term "degree". I do not like index, because it implies a banality--and eschews the fact that this indexing is done off of complexity of the iteration. Where as degree, implies we go up a level similarly to polynomials. This also dates back to older work I had done, where we write:
\[
P(z) = \sum_{j=0}^n a_j \left(\alpha \uparrow^j z\right)\\
\]
Where now \(P\) is of "degree n". Which I wrote as \(\deg(P) = n\). If you apply the super function operator \(\uparrow\), then \(\deg (\uparrow P) = \deg(P) + 1\). Where then, by triviality: \(\deg(\alpha \uparrow^n z) = n\). But I think "degree", might be a tad too general. We don't really say "the degree of \(x^n\) is \(n\)"--despite it being correct, it rolls off the tongue wrong. Instead, we call \(n\) the power of \(x\).
In that regard, I really like "rank". Where the "rank" of the hyper-operator plays the role of the power of \(x\).But when combining different hyper-operators together; we take the "max of the ranks" to get the "degree".
Also, I thought I'd add. I could never fully define this \(\deg\) function, but I could justify its existence very well; and It tides a bit next to "rank" but is an independent idea.
Any function \(f_0\) is a finite combination of \(\alpha \cdot z\), \(\alpha + z\), under composition and addition is \(\deg f = 0\). From this, any function \(f_n = \uparrow f_{n-1}\), where \(\deg(f_n) = n\) is found by taking some \(deg(f_{n-1}) = n-1\) and writing:
\[
f_n(z) = \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{j=0}^\infty f_{n-1}^{\circ j}(z)\frac{w^j}{j!}\\
\]
I could never totally show this \(\deg\) was well defined. But it is well defined for the bounded analytic hyper operators. And it spits out the right answer. Things might get more tricky for higher levels. And it only works as a hierarchy on superpositions and compositions of hyper-operators in the bounded case.
So all in all, I support the term "rank"; though I use it to represent the rank of a hyperoperator; the rank of how deep of a recursion you need; the rank of how many Mellin transforms we have to do. Where then, degree, is a general term to deal with more functions, but still preserving the main tenets of "rank".
The "rank" of a hyperoperator is \(n\) is the same as the power of a monomial \(x^n\) is \(n\). The "degree" of a function \(f\), is the same as the degree of a polynomial \(\sum_{j=0}^n a_j x^j\); where it may have no obvious "rank" (it is not a monomial).
\[
P(z) = \sum_{j=0}^n a_j \left(\alpha \uparrow^j z\right)\\
\]
Where now \(P\) is of "degree n". Which I wrote as \(\deg(P) = n\). If you apply the super function operator \(\uparrow\), then \(\deg (\uparrow P) = \deg(P) + 1\). Where then, by triviality: \(\deg(\alpha \uparrow^n z) = n\). But I think "degree", might be a tad too general. We don't really say "the degree of \(x^n\) is \(n\)"--despite it being correct, it rolls off the tongue wrong. Instead, we call \(n\) the power of \(x\).
In that regard, I really like "rank". Where the "rank" of the hyper-operator plays the role of the power of \(x\).But when combining different hyper-operators together; we take the "max of the ranks" to get the "degree".
Also, I thought I'd add. I could never fully define this \(\deg\) function, but I could justify its existence very well; and It tides a bit next to "rank" but is an independent idea.
Any function \(f_0\) is a finite combination of \(\alpha \cdot z\), \(\alpha + z\), under composition and addition is \(\deg f = 0\). From this, any function \(f_n = \uparrow f_{n-1}\), where \(\deg(f_n) = n\) is found by taking some \(deg(f_{n-1}) = n-1\) and writing:
\[
f_n(z) = \frac{d^z}{dw^z}\Big{|}_{w=0} \sum_{j=0}^\infty f_{n-1}^{\circ j}(z)\frac{w^j}{j!}\\
\]
I could never totally show this \(\deg\) was well defined. But it is well defined for the bounded analytic hyper operators. And it spits out the right answer. Things might get more tricky for higher levels. And it only works as a hierarchy on superpositions and compositions of hyper-operators in the bounded case.
So all in all, I support the term "rank"; though I use it to represent the rank of a hyperoperator; the rank of how deep of a recursion you need; the rank of how many Mellin transforms we have to do. Where then, degree, is a general term to deal with more functions, but still preserving the main tenets of "rank".
The "rank" of a hyperoperator is \(n\) is the same as the power of a monomial \(x^n\) is \(n\). The "degree" of a function \(f\), is the same as the degree of a polynomial \(\sum_{j=0}^n a_j x^j\); where it may have no obvious "rank" (it is not a monomial).