Jabotinsky's iterative logarithm

[bo198214]

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:

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. if f°g = g°f then the operators FM and GM commute.
   
4. if FM and GM commute, their eigenvectors are the same (statement extrapolated from finite matrices)
   
5. so the only different characteristic of FM and GM is the value of their second eigenvalue u^1 and v^1
   
6. but v can be expressed as x'th power of u: v=u^x (or if v is negative and u is positive u=v^y)
   
7. this exponent of the eigenvalue u (resp v) is just the height-parameter, so g=f°x or f=g°y
   

Yes! (However "height" sounds strange when used for iteration of functions which are not exponentials)

[*] remark: this argumentation seems to include, that f and g have the same base-parameter

There is no base. \( f \) and \( g \) are arbitrary analytic functions with development and fixed point at 0.