Iterated function convergence Daniel Long Time Fellow    Posts: 279 Threads: 94 Joined: Aug 2007 12/17/2022, 07:37 PM Fractional iterated function. Note each of the following are combinatorically based - finite inputs result in a finite output. Polynomial, exponential, derivatives, summations, partitions, Bell polynomials, recursive Bell polynomials are all combinatorial. The function to be iterated is homomorphic. So the following fractional iterated function is convergent. $H(0,t)=L$, where $L$ is a fixed point $H(1,t)=f'(L)^t$ the Lyapunov multiplier, denoted $\lambda$, with $\lambda \ne 0$. $H(n,t)=\sum_{r=0}^\infty(\sum_{k=1}^n \frac{f^{(k)}(L)}{k!} B_{n,k}(H(1,t-1),\ldots, H(n-k+1,t-1)))^r$ $f^t(x)=\sum_{n=0}^\infty\frac{1}{k!} H(k,t) (x - L)^k$ Daniel JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 12/18/2022, 01:40 AM (This post was last modified: 12/18/2022, 04:40 AM by JmsNxn.) (12/17/2022, 07:37 PM)Daniel Wrote: Fractional iterated function. Note each of the following are combinatorically based - finite inputs result in a finite output. Polynomial, exponential, derivatives, summations, partitions, Bell polynomials, recursive Bell polynomials are all combinatorial. The function to be iterated is homomorphic. So the following fractional iterated function is convergent. $H(0,t)=L$, where $L$ is a fixed point $H(1,t)=f'(L)^t$ the Lyapunov multiplier, denoted $\lambda$, with $\lambda \ne 0$. $H(n,t)=\sum_{r=0}^\infty(\sum_{k=1}^n \frac{f^{(k)}(L)}{k!} B_{n,k}(H(1,t-1),\ldots, H(n-k+1,t-1)))^r$ $f^t(x)=\sum_{n=0}^\infty\frac{1}{k!} H(k,t) (x - L)^k$ I'm very fascinated by this Daniel, but also a tad skeptical. What is your reasoning for: $H(n,t)$ Is convergent. I largely don't doubt you, but I also doubt this holds entirely. For example, set $L=0$, and $f'(0) = 1$. Then we know for a fact that: $f^{\circ 1/2}(x) = \sum_{n=0}^\infty H(n,1/2) \frac{x^n}{n!}\\$ DOES NOT CONVERGE. This is because $|H(n,1/2)| >n!^2\delta$ for some $0 < \delta$. Which forces divergence of this series. But this makes an accurate asymptotic series, which can "look holomorphic", but isn't actually. I absolutely believe this converges for $|\lambda| \neq 0,1$; but I think it'll fail for some values on the unit circle; namely roots of unity. So if $\lambda =e^{2 \pi i \frac{p}{q}}$ for $p,q \in \mathbb{Z}$, then this will absolutely not converge. Baker proved this unconditionally. For $\lambda = e^{2\pi i \xi}$ for $\xi \in \mathbb{R}/\mathbb{Q}$, we do have an iterate, and I'd be willing to put money on this converging. For $\lambda =0$, we might get lucky with an asymptotic series, but I doubt it converges, as the iteration has a branch cut spawning at $L$, and therefore there's no convergent Taylor series, but they're might be an asymptotic series. Finding $f^{\circ 1/2}$ equates to evaluating $z^{\sqrt{\ell}}$, where $f(z) = z^{\ell} + O(z^{\ell+1})$, which has a branch cut at $0$. I hope you don't think I'm trying to dissuade you Daniel. I believe that everything you've written is at least an Asymptotic Expansion (Which is still super fuckin valuable). And for the most part it is a holomorphic expansion; with just some problem points here and there. I also really like this expansion because it's incredibly clean cut EDIT: I know your work largely agrees with Gottfried; but I have not seen if you've double checked this. What do your coefficients: $H(n,1/2)\\$ Look like, when $f(z) = e^{z} - 1$. Whereby we write: $f^{\circ 1/2}(z) = \sum_{n=1}^\infty H(n,1/2) \frac{z^n}{n!}\\$ Gottfried chose these coefficients as $g_n = H(n,1/2)/n!$--whereby a fairly large result we have pulled out of the woodwork is that $g_n \sim c^nn!$--for some $c > 0$. And we found this dates back nearly a hundred years to scattered papers by Baker, and later, Ecalle. Who showed that necessarily your taylor series does not converge. Instead it's an asymptotic series (it looks so close to converging, and we can approximate indefinitely, but the summed series is divergent). Additionally, this may only be noticed at 100 digits, or so; so even though it can be directly accurate for 20 digits is exemplary of how accurate the asymptotic series is. But it is not a Taylor series. It is an Euler series $\sum_k k!z^k$, which is a really really accurate asymptotic series--which Borel made his name on, justifying Euler's technique to prove this is an analytic function. I'm just trying to point this out, because I believe you are falling into the trap that $\sum_k k!z^k$ is holomorphic--when it isn't. It is holomorphic near $z=0$; but not at $z=0$. The sum: $E(z) = \sum_{k=0}^\infty k!z^k\\$ Is holomorphic for $\mathbb{C}/[0,\infty)$. But this series is divergent. We have to write this as: $E(z) = \int_0^\infty \frac{e^{-t}}{1-zt}\,dt = \sum_{k=0}^\infty z^k \int_0^\infty e^{-t}t^k\,dt = \sum_{k=0}^\infty k!z^k\\$ Which FORMALLY equals the above series... Iconically this integral converges uniformly for $z \in \mathbb{C}/[0,\infty)$. So that $E(z)$ is holomorphic for $\mathbb{C}/[0,\infty)$. Which is the exact same thing we get with $f^{\circ 1/2}$. And at $z=0$ it's so well behaved we get an Euler series.... But if you see: $E(-0.1) \approx \sum_{k=0}^{8} k!(-0.1)^k\\$ So that the partial sums $\sum_{k=0}^Nk!z^n \approx E(z)$--for$z\in \mathbb{C}/[0,\infty)$ and $|z|<\delta$ where $N = N(\delta)$... What happens is we can use the values $a_n = H(n,1/2) = O(n!^2)$, to essentially "reconstruct" the holomorphic function (using Euler's work on this series--but justifying it with the rigor of later mathematicians). But your series is divergent; but not all is lost. Which is seen by your accurate evaluations. Which I don't doubt are very good. But try doing it for 100-1000 digits, and you'll start to see it fail We need some extra magic for these cases. Despite being hyper accurate approximations, things start to deteriorate at certain levels. And we have to double check. Especially when it comes to HOLOMORPHY/ANALYTIC! « Next Oldest | Next Newest »

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