Edit: Changed \( O(n) \) to \( O(z^n) \).
I primarily focus on iterated functions instead of tetration because a general theory of complexly iterated functions serves as a strong foundation for researching the hyperoperators. I study complex tetration, but not real tetration because I don't believe that it can pass the following test. Please let me know if I am wrong.
Let
\( f(0)=0, f^0(z)=z, f^1(z)=f(z) \)
with
\( a,b,z\in \mathbb{C} \) and
\( n\in \mathbb{Z^+} \) where the Taylor series for \( f(z) \) obeys
\( f^a(f^b(z)) - f^{(a+b)}(z) = O(z^n) \).
I consider this the gold standard in extending the iteration of functions from the natural numbers.
I have derived and proven symbolically using Mathematica that the Taylors series for \( f(z) \) satisfies
\( f^a(f^b(z)) - f^{(a+b)}(z) = O(z^{19}) \)
and that iterated sin function
\( \sin^a(\sin^b(z)) - sin^{(a+b)}(z) = O(z^{39}) \)
Has anyone else done this sort of consistency test? If so, what were your results for \( o(z^n) \)?
I primarily focus on iterated functions instead of tetration because a general theory of complexly iterated functions serves as a strong foundation for researching the hyperoperators. I study complex tetration, but not real tetration because I don't believe that it can pass the following test. Please let me know if I am wrong.
Let
\( f(0)=0, f^0(z)=z, f^1(z)=f(z) \)
with
\( a,b,z\in \mathbb{C} \) and
\( n\in \mathbb{Z^+} \) where the Taylor series for \( f(z) \) obeys
\( f^a(f^b(z)) - f^{(a+b)}(z) = O(z^n) \).
I consider this the gold standard in extending the iteration of functions from the natural numbers.
I have derived and proven symbolically using Mathematica that the Taylors series for \( f(z) \) satisfies
\( f^a(f^b(z)) - f^{(a+b)}(z) = O(z^{19}) \)
and that iterated sin function
\( \sin^a(\sin^b(z)) - sin^{(a+b)}(z) = O(z^{39}) \)
Has anyone else done this sort of consistency test? If so, what were your results for \( o(z^n) \)?
Daniel