Some basic thoughts
If we want to summarize our operation sequence [n] we can perhaps write:
\( b[n]x=f_{n,b}(x) \), where
\( f_{1,b}(x) = b+x \)
\( f_{n+1,b} = t\to f_{n,b}^{\circ t}(1) \), \( n\ge 1 \)
So we have a certain operator \( E \) that assigns to a given function \( f \) the function \( E(f)=t\to f^{\circ t}(1) \), this operator may be based on the natural Abel method, the diagonalization method, or the regular Abel method (with restrictions of \( b \)) and we can write
\( f_{n+1,b} = E(f_{n,b}) \), moreover
\( f_{n+1,b} = E^{\circ n}(f_{1,b}) = E^{\circ n}(x\mapsto b+x) \)
\( b[n+1]x = E^{\circ n}(f_{1,b})(x) \).
Note: I original found it more appropriate to start with the addition as 0th instead of the 1st operation. However I adapted to the already established nomenclature. If we would stick to my original counting (addition as 0th operation) we had the better looking formula \( b[n]x=E^{\circ n}(f_{0,b})(x) \)
As we have some methods for real functions to switch from \( f^{\circ n} \) to non-natural iterations \( f^{\circ t} \) maybe there are also methods for the operator \( E \) to compute \( E^{\circ t} \) which then gives definition for real and complex iteration ranks:
\( b[t]x=E^{\circ t-1}(f_{1,b})(x) \).
In the moment however it is even unclear how to generally express such operators which map powerseries.
At least we can determine that \( E^{-1} \) is the inverse operator, i.e. \( E^{-1}(g) \) is a function \( f \) such that \( g(x+1)=f(g(x)) \), i.e. \( E^{-1}(g)(x)=g(g^{-1}(x)+1) \). So \( E^{-1} \) is unique (independent of the method of \( E \) and independent on the initial condition).
If we compute \( f_{0,b} \) from this view point we get also:
\( f_{0,b}(x)=E^{-1}(f_{1,b})(x)=f_{1,b}(f_{1,b}^{-1}(x)+1)=b+(x-b+1)=x+1 \) and
\( f_{-1,b}(x)=E^{-1}(f_{0,b})(x)=f_{0,b}(f_{0,b}^{-1}(x)+1)=(x-1+1)+1=x+1 \)
...
If we want to summarize our operation sequence [n] we can perhaps write:
\( b[n]x=f_{n,b}(x) \), where
\( f_{1,b}(x) = b+x \)
\( f_{n+1,b} = t\to f_{n,b}^{\circ t}(1) \), \( n\ge 1 \)
So we have a certain operator \( E \) that assigns to a given function \( f \) the function \( E(f)=t\to f^{\circ t}(1) \), this operator may be based on the natural Abel method, the diagonalization method, or the regular Abel method (with restrictions of \( b \)) and we can write
\( f_{n+1,b} = E(f_{n,b}) \), moreover
\( f_{n+1,b} = E^{\circ n}(f_{1,b}) = E^{\circ n}(x\mapsto b+x) \)
\( b[n+1]x = E^{\circ n}(f_{1,b})(x) \).
Note: I original found it more appropriate to start with the addition as 0th instead of the 1st operation. However I adapted to the already established nomenclature. If we would stick to my original counting (addition as 0th operation) we had the better looking formula \( b[n]x=E^{\circ n}(f_{0,b})(x) \)
As we have some methods for real functions to switch from \( f^{\circ n} \) to non-natural iterations \( f^{\circ t} \) maybe there are also methods for the operator \( E \) to compute \( E^{\circ t} \) which then gives definition for real and complex iteration ranks:
\( b[t]x=E^{\circ t-1}(f_{1,b})(x) \).
In the moment however it is even unclear how to generally express such operators which map powerseries.
At least we can determine that \( E^{-1} \) is the inverse operator, i.e. \( E^{-1}(g) \) is a function \( f \) such that \( g(x+1)=f(g(x)) \), i.e. \( E^{-1}(g)(x)=g(g^{-1}(x)+1) \). So \( E^{-1} \) is unique (independent of the method of \( E \) and independent on the initial condition).
If we compute \( f_{0,b} \) from this view point we get also:
\( f_{0,b}(x)=E^{-1}(f_{1,b})(x)=f_{1,b}(f_{1,b}^{-1}(x)+1)=b+(x-b+1)=x+1 \) and
\( f_{-1,b}(x)=E^{-1}(f_{0,b})(x)=f_{0,b}(f_{0,b}^{-1}(x)+1)=(x-1+1)+1=x+1 \)
...
