generalizing the problem of fractional analytic Ackermann functions

[JmsNxn]

That's a good way to approach the question. I'm not too familiar with carleman matrices but I think it goes something like this

(...)

Hi James -

unfortunately I can't follow in the above. But concerning tetration of carleman-matrices themselves there is a simple example: the pascal-matrix is the carleman-matrix for the operation of incrementation by 1 and powers of it are carleman-matrices for the operation of addition. I've worked out an example how to tetrate the pascal-matrix, perhaps this is useful (and possibly generalizable). See http://go.helms-net.de/math/tetdocs/index.htm go to the short statement at "pascal matrix tetrated" and open http://go.helms-net.de/math/tetdocs/Pasc...trated.pdf

Nearly all what I'm doing in tetration is based on this concept of "carleman-matrices" and I might be able to answer if you have some concrete questions. I did not know that name when I stumbled on that concept so my first expositions of my fiddlings are mostly in threads headed by "matrix-method" and might be informative/helpful before (though it is extremely exploratory and not well structured from the beginning)

Gottfried

Thanks Gottfried, I'll take a look. It may be in my interest to research more into how the matrix method actually works.

It seems like the matrix method may have a potential solution. I stress the may.


So far, from what I can make of it, it would require taking a limit, and an exponential continuum sum.

In the sense \( M \) is a carlemann matrix.

\( T[M]_{n=0}^{R}\,\, g(n) = ((...(M[f]^{g(0)})^{g(1)})^{g(2)})...)^{g( R )} \)

this isn't really tetration, but rather a type of left handed tetration. This would require a continuum sum type object for fractional R iff the following identity isn't true:

\( (M[f]^a)^b = M[f]^{a \cdot b} \)

which I think is given if matrix multiplication isn't commutative. However, this seems to contradict how Carlemann matrices work, because evidently

\( M[f^{\circ t}(x)] = M[(f^{\circ \frac{t}{2}})^{\circ 2}(x)] = M[f^{\circ \frac{t}{2}}(x)]^2 = (M[f(x)]^{\frac{t}{2}})^2 \)

but we already know

\( M[f^{\circ t}(x)] = M[f(x)]^t \)

so this would imply

\( (M[f]^a)^b = M[f]^{a \cdot b} \)



I'm a little bit wishy washy now.

But I think in general I just need to better understand most fractional iteration methods. I doubt this type of reasoning is unique to Carlemann Matrices. And to my knowledge, the matrix method doesn't quite make the cut--I don't remember why, though; does it fail for \( b < e^{\frac{1}{e}} \)?


Maybe if I could make sense of Kouznetsov's method of finding superfunctions I might be able to iterate that.



And if your confused, you shouldn't be, this is fairly simple.

It's following this reasoning, if:

\( F\{ f \}(x) \) is the super function of f(x)

and
\( F\{ F \{ f \} \} (x) \) is the super function of \( F \{ f \}(x) \) and the second superfunction of f(x)

so that in general

\( F^{t}\{ f \}(x) \) is the t'th super function of f(x)

what is the value of
\( F^{\frac{1}{2}} \{ f \} (x) \)


There are plenty of ways to find the super function of f and to define \( F \{f \} (x) \) So I'll try to experiment with as many as possible. There should be one that sticks out though, or is simplest. Carlemann matrices were just the first.