Jabotinsky's iterative logarithm

[Gottfried]

But take into account that Jabotinsky is only considering functions with f(0)=0, i.e. which have a fixed point at 0. Generally all formal powerseries co mputations are restricted to this case because otherwise the coefficients of composition are no finite expressions (in terms of power series coefficients) anymore.

Ah -yes. If he matrix-operator is triangular, then the diagonal is the sequence of consecutive powers of f'(0)/1! . So, in matrix-lingo it is the statement:[updated]

1. assume functions f and g having their associated matrix-operators FM and GM triangular (f(0),g(0)=0)
   
2. Then both operators have sets of eigenvalues, which consist of the consecutive powers of base-parameter, say u for FM and v for GM
   
3. f and g may be seen as iterates of f0°a resp g0°b. The second eigenvalue of FM is then u0^a and of GM is v0^b
   
4. if f°g = g°f then the operators FM and GM commute.
   
5. if FM and GM commute, their eigenvectors are the same (statement extrapolated from finite matrices)
   
6. the eigenvectors consists of polynomials in u0 and v0.
   Proposal: from the polynomial composition of coefficients in the eigenvectors it follows, that u0=v0
   
7. from u0=v0 it follows, that v = v0^b = u0^b =u^(b/a) or u=u0^a = v0^a = v^(a/b)
   
8. from this follows, that also f0=g0 and
   
9. f=f0°a =g0°a =g°(-b+a)
   or g=g0°b = f0°b = f°(-a+b) ,
   so f and g are iterates of each other
   

Hmm. this statement is worth to be put into the matrix-facts-library (to be created)...

Gottfried