(07/14/2022, 02:45 PM)MphLee Wrote: Excuse me Daniel... let \(f:X\to X\) be a function, I'd call an iteration of \(f\) an action of an abelian group \(A\) on \(X\) s.t. for some \(u\in A\) we have \(f_u=f\). In other words, when we talk about iteration I believe that it goes without saying that we assume the law \[f_{a+b}=f_a\circ f_b\] to hold on the nose and not only asymptotically.
So my question is: why don't you ask your solutions to satisfy \(f_{a+b}=f_a\circ f_b\) directly instead of \(f_{a+b}-(f_a\circ f_b) \in \mathcal O({\rm id}^k)\)?
I'm totally newbye in this so I understand like 1% of this thread... maybe Daniel, if you could be a bit more detailed on your assumptions and "story-tell" it a bit it may be easier for non-specialists like me...
See Bell Polynomials of Iterated Functions for a complete treatment of the following material.
The following is the Taylor's series of \(f^t(z)\) with a non-superattracting fixed point translated to zero.
\(f^t(z)=f'(0)^t z + \left( f''(0) \sum_{k_1=0}^{(t-1)} f'(0)^{2t-k_1-2} \right) z^2 + \left( f'''(0) \sum_{k_1=0}^{(t-1)}f'(0)^{3t-2k_1-3}+3f''(0)^2 \sum_{k_1=0}^{(t-1)} \sum_{k_2=0}^{(t-k_1-2)} f'(0)^{3t-2k_1-k_2-5} \right) z^3+\mathcal{O}(z^4) \)
Computer algebra validates \(f^{a+b}(z)-f^a(f^b(z))=\mathcal{O}(z^k)\) out to \(\mathcal{O}(z^9)\). This sounds meager until you consider that the coefficient for \(z^8\) consists of 660,032 expressions where each expression can be an 8\(^\text{th}\) order polynomial.
Setting \(f(z)=\sin(z)\) results in
\(\sin^t(z)=z-\frac{t}{6}z^3 + \left( -\frac{t}{30}+\frac{t^2}{24}\right) z^5 + \left( -\frac{41t}{3780}+\frac{t^2}{45}-\frac{5t^3}{432}\right)z^7 + \left( -\frac{4t}{945}+\frac{67t^2}{5670}-\frac{71t^3}{6480}+\frac{35t^4}{10368}\right)z^9+\mathcal{O}(z^{11})\)
See iterated sin function for more information including a very short Mathematica program that computes the coefficients of the iterated sin function.
Why not prove \(\sin^{a+b}(z)=\sin^a(\sin^b(z))\) instead of \(\sin^{a+b}(z)-\sin^a(\sin^b(z))=\mathcal{O}(z^k)\)? Mathematica has a difficult time of comparing two different complex expressions for equality, but by using the version that used subtraction and repeated simplification, terms repeatedly cancel out until we are just left with a big O expression.
Why do I care? Because I am guarded against two things, the movement between use of Schroder's and Abel's functional equations; that topological conjugacy is understood and respected. Then I recently read Schroeder's Equation and the connection between Schroeder's and Abel's equation. My second concern is that proofs based on Kneser's paper may well begin satisfying \(f^{a+b}(z)=f^a(f^b(z))\), but I question whether the identity survives the mapping to the unit circle and then into the real line.
Daniel