Ivars Wrote:\( \phi(f^{\circ t}(x))=\phi(x)+t \).
e.g.
\( \phi(f^{\circ I}(x))=\phi(x)+I \).
If we turn it upside down, we can may be use the result of t iterations of a function to define what COUNTING with t (e.g. t=I) means.
We may get functions which are only COUNTABLE in certain ways, so that Abel equation is not possible to solve for all t.
Would that make sense?
Ivars
Just some speculation:
if we assume the divergent summation 1+3+9+...+3^k+... = 1/(1-3) = -1/2 applicable everywhere in the context of powerseries, then we may also assume
x^(1+3+9+...) = x*x^3*x^9*... = x^(-1/2)
and for iteration
...f°9(f°3(f(x))) = f°(1+3+9+...)(x) = f°(-1/2)(x)
:cool:
Then we may take another series in the iterator; I don't remember at the moment which, but I think, there are some divergent series of rational or even integer summands, for whose value the replacement by I makes sense/is consistent. Then we had the same way
... f°c(f°b(f°a(x))) = f°(a+b+c+...)(x) = f°I(x)
But don't know, whether this makes sense ...
Gottfried Helms, Kassel