08/12/2022, 05:26 AM
(08/12/2022, 01:28 AM)tommy1729 Wrote: is there a related thread ?
...
tommy1729
The first one followsÂ
https://math.eretrandre.org/tetrationfor...p?tid=1318
on page 4 & 5 me and james discussed about the generalized iterations that has such fixed point and multiplier which is "non-constructable", for example \(f(z)=-z+z^2\), it's almost impossible to generate \(f^{\frac{1}{2}}(z)\) from the fixed point z=0 because its multiplier is -1 in the form \(\lambda=e^{2\pi i\frac{1}{2}}\), but can still generate the half iterate at z=2. It matters because if it's true then we can always generate any iterations of any such functions as we want, otherwise it'll leave us with tons of heavy arduous computative problems. You can try to build such function \(f^{\frac{1}{2}}(z)=\pm iz+O(z^2)\) and may won't succeed.
The second is a generalized idea, for example hyperoperations.
We define S[f](z) or S as an operator whose results are the superfunction families of f, then hyperoperations can be expressed as a branch cut of \(S^n[f](z)\), precisely \(a[n]b\in{S^n[a[0]b]}\) wrt b.
And also, if we define a S operator for S operator, or let's call it T: \(T[S[f](z)](z+1)=S[T[S[f](z)](z)](z)\), then we have hyperhyperoperators (idk the exact name lol but such concept exists widely especially in googleplex-ology(?))
And more, for example any functional equation can be seen as \(H[f(z)](z)=0\), H is an operator, thus if we find exp(H) and log(H), we get immediately the solution by \(f(z)=H^{-1}[0](z)=e^{-\log(H)}[0](z)\)
Regards, Leo

