generalizing the problem of fractional analytic Ackermann functions

[JmsNxn]

I just got the feeling that you were implying \( f^{\diamond n}(x) \) is inconsistent seperate from the diamond operator; which is not true. It's very simple and nothing about it implies inconsistency, unless you think an infinite sequence of superfunctions is inconsistent.

hmm ...

how does a converging infinite sequence of uniquely defined superfunctions look like any way ?

fractal ??

another question or remark ..

if a superfunction look like ( carleman f(z) ) ^ x , does that imply that fractional superfunction operators look like ( carleman f(z) ) ^ ^ x ?

if so , we need to extend the bases E [1,eta]U[sqrt(e),oo[ to matrix bases ?

does this imply that the (carleman) matrices require |det| = E[1,eta]U[sqrt(e),oo[ ??

again - as usual - many questions , more questions than answers.

regards

tommy1729

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


Beware: the following maybe extremely inconsistent... Enter at own discretion


\( M[(f \circ g) (x)] = M[f(x)]M[g(x)] \)

and by the recurrence relation

\( f^{\diamond n+1}(x) = (f^{\diamond n})^{\circ x} ( C_n ) \)

for any appropriate constant \( f^{\diamond n+1}(0) = C_n \) that does not produce singularities or other problems.

therefore

\( M[f^{\diamond 1}(x)] = M[f^{\circ x} ( C_0 )] = M[f( C_0 )]^x \)

which is the traditional superfunction equation.

continuing this we would get

\( M[f^{\diamond n + 1} ( x ) ] = M[(f^{\diamond n})^{\circ x}(C_n)] = M[f^{\diamond n} ( C_n )]^x \)

\( M[f^{\diamond n+1}(x)] = M[(f^{\diamond n-1})^{\circ C_{n}} ( C_{n-1} ) ]^x = (M[f^{\diamond n-1} ( C_{n-1} ) ]^{ C_n })^x \)

which would give:

\( M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x \)

This isn't good though, this would imply the matrix method doesn't work for the Ackermann function since \( C_k = 1 \)

Arghhhh if only I knew more about matrices. I hope you can help me out.

Hopefully your guess is right, that it somehow turns into tetration across matrices. At least that way we get something non-contradictory.


but there is hope, if

\( M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x \)

is valid then we can have a solution for the modified Ackermann function; namely

\( a\,\,\bigtriangleup_\sigma^K\,\,b = a\,\,\bigtriangleup_{\sigma - 1}^K\,\,(a\,\bigtriangleup_\sigma^K\,\,b-1) \)

where

\( a\,\,\bigtriangleup_{1}^K\,\,b = a + b \)

and

\( a\,\,\bigtriangleup_{n}^K\,\,K = a \) for \( n \in \mathbb{Z};\,\,n > 1 \)


for multiplication

\( a\,\,\bigtriangleup_{2}^K\,\,b = a \cdot (b - K + 1) \)

and for exponentiation

\( a\,\,\bigtriangleup_{3}^K\,\,b + K = a^{b+1} + (1-K) \cdot \sum_{c = 1}^{b} a^c \)

\( a\,\,\bigtriangleup_{3}^K\,\,b = a^{b-K+1} + (1-K) \cdot \sum_{c = 1}^{b-K} a^c \)

tetration would be ridiculously complex.

with this we have the relation

\( f^{\diamond \sigma + 1}(b) = a\,\,\bigtriangleup_{\sigma + 1}^K\,\,b = ((...(M[f(a)])^{f^{\diamond 1}(0)})^{f^{\diamond 2}(0)})...)^{f^{\diamond \sigma}(0)})^b \)

with \( f(x) = x + 1 \) and \( f(a) = f^{\diamond 1}(1) \)

Hopefully you or somebody can correct me if I'm wrong. I think that I'm wrong; this seems too contradictory and fishy to be right... The fact that it fails for the usual Ackermann function makes me angry D:<

If this is right though, we may have a tool for solving half-superfunctions given certain parameters that \( f^{\diamond n}(0) \neq 1 \) which is the equivalent of \( f^{\diamond n}(x) \neq (f^{\diamond n - 1})^{\circ x}(1) \)

In a sense you were right about the tetration intuition. Except that it's like a left-handed tetration! which forms an independent operator over matrices since matrix multiplication is not commutative!

EDIT: The formula seems to change from the original one

\( M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x = ((...(M[f^{\diamond 1}(1)]^{f^{\diamond 2}(0)})^{f^{\diamond 3}(0)})...)^{f^{\diamond n+1}(0)})^x \)  

to

\( ((...(M[f^{\diamond 1}(1)]^{f^{\diamond 2}(0)})^{f^{\diamond 3}(0)})...)^{f^{\diamond n}(0)})^x \)

not sure why that is, must've made a minor mistake somewhere...

Too tired right now to figure out where. I think I've slashed through enough mathematical jungle for one day.

Maybe you can clarify my errors...