07/12/2022, 05:27 AM
(07/12/2022, 03:05 AM)Daniel Wrote:(07/03/2022, 01:46 PM)Daniel Wrote: ...
My ultimate sanity test is to prove symbolically that using the Taylor's series for \( f^n(z) \) that \( f^{a+b}(z)-f^{a}(f^{b}(z))=\mathcal{O}(z^k) \). For my check I was able to get to \( \mathcal{O}(z^{29}) \) where the Lyapunov multiplier \( \lambda \) is neither zero or a root of unity and the origin is set to a fixed point not infinity. I used no floating point in my calculations, only rational numbers, so I could obtain an exact answer.
...
No one provided an answer to my question, so I will come at it from another direction. What software that members have written that is based on rational numbers and not floating point?
Very difficult question to answer, Daniel. And I think I get the question more clearly now.
You want to approach the answer from \(\mathbb{Q}[z]\); the space of polynomials with rational coefficients \(p_m(z)\). This means, you are asking to look at iterations like:
\[
f^n(z) = \lim_{m\to\infty} p_m(z)
\]
In which, you are choosing the minimal rational polynomial near this solution.
I apologize if I'm skipping something, but does this sound what you're getting at?
If so, I have no record of anyone ever doing something like that. That is beyond fascinating. I've never seen anyone approach the question like that.
I don't see why you'd use that, but that's super cool!
It makes sense though, and I whole heartedly agree with you that the algorithm converges. Choosing rational polynomials, constructs the same limit. But choosing rational coefficients gets the number theorist part of me going, lol.
Regards