01/19/2023, 06:31 PM

Hey guys it's been so long time and things in uni are getting busier than ever

Personally I'd expect myself to pass GRE and learn abroad, most potentially in USA

And I postponed so many researches but except for the kinda conjecture that I posted earlier, that in the sense of asymptotic expansion, every function and its iterations can form a group under composition, and thus we can compute the asymp expansion of the iterations in so much convenience.

I just gave a try to, still the old equation friend, \(f(f(z))=-z+z^2\), at \(z=0\) with a multiplier \(-1\) as fuck, it's trivially proved that there's just no Taylor Expansion for this function, I kinda got stucked in here, since I used to consider the iterative Taylor Expansions(as asymtotic expansion) as a group under composition in the sense aforementioned, and by this equation you can prove that they cannot form a group, but only excluded the nonconstructible multipliers(amplifiers).

James was so correct, the petals do generate iterates, the point I got stuck was how to compute, substantially.

If you do care about multivalued-ness, it can indeed construct an Abel function of such functions, yet still arduous and too skillful to approach a well-behaved, not nonsensical inverse of that Abel function, and hence a superfunction.

But it occurs to me that the wider asymp expansions(still conjecture though), in the multivalued sense, form a group. So I tried to grow a computational method to approach any half iterate of a function with nonconstructible multipliers, namely \(-1\).

If we assume \(\log(az)=\log(a)+\log(z)\) for small \(z\) and all \(a\), and bring the main branch of \(\log(z)\) in the evaluation, we have (I took only one branch)

$$f(z)=-iz-\frac{1-i}{2}z^2+z^3\big(\frac{1}{2}+\frac{2\log(z)}{\pi}\big)-iz^4\big(\frac{i-1}{2}-\frac{i}{\pi}+\frac{(2-3i)\log(z)}{\pi}\big)$$ $$-iz^5\frac{1}{8\pi^2}\big(\pi ((10i-8 )+(5-6i)\pi)+4i(4i+(3+8i)\pi)\log(z)-48\log(z)^2\big)+O(z^6\log(z)^3)$$

I can't be more excited, you can check how well \(f(f(z))\) approximates \(-z+z^2\) at least in a small region around z=0 and missing accuracy in some directions in the complex plane. Here's rough Re-Im plot showing \(f(f(z))\) (undoubtedly orange) and \(-z+z^2\) (blue). (Dotted line = Im, Solid line = Re)

But "unfortunately" the existence of this expansion might declare the expiration of conjectures that we can approx it by limits such like:

\(f_c(z)=\displaystyle\lim_{c\to -1}{cz+z^2}\) and so derived methods.

Personally I'd expect myself to pass GRE and learn abroad, most potentially in USA

And I postponed so many researches but except for the kinda conjecture that I posted earlier, that in the sense of asymptotic expansion, every function and its iterations can form a group under composition, and thus we can compute the asymp expansion of the iterations in so much convenience.

I just gave a try to, still the old equation friend, \(f(f(z))=-z+z^2\), at \(z=0\) with a multiplier \(-1\) as fuck, it's trivially proved that there's just no Taylor Expansion for this function, I kinda got stucked in here, since I used to consider the iterative Taylor Expansions(as asymtotic expansion) as a group under composition in the sense aforementioned, and by this equation you can prove that they cannot form a group, but only excluded the nonconstructible multipliers(amplifiers).

James was so correct, the petals do generate iterates, the point I got stuck was how to compute, substantially.

If you do care about multivalued-ness, it can indeed construct an Abel function of such functions, yet still arduous and too skillful to approach a well-behaved, not nonsensical inverse of that Abel function, and hence a superfunction.

But it occurs to me that the wider asymp expansions(still conjecture though), in the multivalued sense, form a group. So I tried to grow a computational method to approach any half iterate of a function with nonconstructible multipliers, namely \(-1\).

If we assume \(\log(az)=\log(a)+\log(z)\) for small \(z\) and all \(a\), and bring the main branch of \(\log(z)\) in the evaluation, we have (I took only one branch)

$$f(z)=-iz-\frac{1-i}{2}z^2+z^3\big(\frac{1}{2}+\frac{2\log(z)}{\pi}\big)-iz^4\big(\frac{i-1}{2}-\frac{i}{\pi}+\frac{(2-3i)\log(z)}{\pi}\big)$$ $$-iz^5\frac{1}{8\pi^2}\big(\pi ((10i-8 )+(5-6i)\pi)+4i(4i+(3+8i)\pi)\log(z)-48\log(z)^2\big)+O(z^6\log(z)^3)$$

I can't be more excited, you can check how well \(f(f(z))\) approximates \(-z+z^2\) at least in a small region around z=0 and missing accuracy in some directions in the complex plane. Here's rough Re-Im plot showing \(f(f(z))\) (undoubtedly orange) and \(-z+z^2\) (blue). (Dotted line = Im, Solid line = Re)

But "unfortunately" the existence of this expansion might declare the expiration of conjectures that we can approx it by limits such like:

\(f_c(z)=\displaystyle\lim_{c\to -1}{cz+z^2}\) and so derived methods.

Regards, Leo