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 Half-iterate exp(z)-1: hypothese on growth of coefficients JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/19/2022, 06:17 AM (This post was last modified: 08/19/2022, 06:18 AM by JmsNxn.) (08/19/2022, 05:54 AM)Leo.W Wrote: (08/19/2022, 05:35 AM)JmsNxn Wrote: Leo, If a function is holomorphic in a neighborhood of $$0$$, then it is expandable in a taylor series at $$0$$. If a functions IS NOT holomorphic in a neighborhood of $$0$$, then it IS NOT expandable in a taylor series at $$0$$. What Baker is saying, is that there is a divergent series at $$0$$--THEREFORE, it is not holomorphic in a neighborhood of $$0$$. (*) Unless it is an LFT-- by which it wouldn't be a Euclidean mapping But James, holomorphic refers to differentiable, not the convergent Taylor series, it can have divergent series, as mostly hypergeometric functions do? Here's another example, consider $$f(z)=\sum_{n\ge0}{_2\mathrm{F}_0(n,a;;c)z^n}$$, this function diverges because all its coefficients refers to a divergent series, $$_2\mathrm{F}_0(n,a;;c)=\sum_{k\ge0}{\frac{(a)_k(n)_k}{k!}c^n}$$, but still can be continued by $$f(z)=-cze^{\frac{z-1}{c}}\big(\frac{z-1}{c}\big)^{a-1}\Gamma(1-a,\frac{z-1}{c})$$, choosing proper a, the function is holomorphic, has a divergent series in some sense but indeed a convergent series(Borel sum). Quote:Ps. what if you don't consider the petals of log⁡(1+z)? because e^z−1 maps Re(z)<−M for very large M into the neighborhood of 0, that is why I defined the branch cuts, so that taking both petals of log⁡(1+z) and a "petal" strip from Re(z)<0∧|Im(z)|<1 produces a holomorphic halfiterate at 0 I wonder if you'd try this? LEO! Have you not heard of Cauchy's theorem. If a function is complex differentiable, it is expandable in a taylor series. If a taylor series converges, it is complex differentiable!!!! HOLOMORPHY = TAYLOR SERIES EXPANDABLE Leo.W Fellow   Posts: 79 Threads: 6 Joined: Apr 2021 08/19/2022, 06:54 AM (08/19/2022, 06:17 AM)JmsNxn Wrote: (08/19/2022, 05:54 AM)Leo.W Wrote: (08/19/2022, 05:35 AM)JmsNxn Wrote: Leo, If a function is holomorphic in a neighborhood of $$0$$, then it is expandable in a taylor series at $$0$$. If a functions IS NOT holomorphic in a neighborhood of $$0$$, then it IS NOT expandable in a taylor series at $$0$$. What Baker is saying, is that there is a divergent series at $$0$$--THEREFORE, it is not holomorphic in a neighborhood of $$0$$. (*) Unless it is an LFT-- by which it wouldn't be a Euclidean mapping But James, holomorphic refers to differentiable, not the convergent Taylor series, it can have divergent series, as mostly hypergeometric functions do? Here's another example, consider $$f(z)=\sum_{n\ge0}{_2\mathrm{F}_0(n,a;;c)z^n}$$, this function diverges because all its coefficients refers to a divergent series, $$_2\mathrm{F}_0(n,a;;c)=\sum_{k\ge0}{\frac{(a)_k(n)_k}{k!}c^n}$$, but still can be continued by $$f(z)=-cze^{\frac{z-1}{c}}\big(\frac{z-1}{c}\big)^{a-1}\Gamma(1-a,\frac{z-1}{c})$$, choosing proper a, the function is holomorphic, has a divergent series in some sense but indeed a convergent series(Borel sum). Quote:Ps. what if you don't consider the petals of log⁡(1+z)? because e^z−1 maps Re(z)<−M for very large M into the neighborhood of 0, that is why I defined the branch cuts, so that taking both petals of log⁡(1+z) and a "petal" strip from Re(z)<0∧|Im(z)|<1 produces a holomorphic halfiterate at 0 I wonder if you'd try this? LEO! Have you not heard of Cauchy's theorem. If a function is complex differentiable, it is expandable in a taylor series. If a taylor series converges, it is complex differentiable!!!! HOLOMORPHY = TAYLOR SERIES EXPANDABLESure I had! I get your point, But does expandable = have a positive radius of convergence? What I've been taught is that there can be 0 radius of convergence(as the finite sum always differentiable?)... and by inversion of Borel sum, any series with $$a_n>O(n!^{-k})$$ can't converge... So this is my pov, by borel sum it must converge, and by multivaluedness and branch cut, it's probable to find a branch holomorphic at z=0... I apologize, I'm stiff and my words stubborn. Anyway I'll say you're right. Please try the construction with a strip branch cut, by the petals of $$e^z-1$$ is different from that by the petals of $$\log(z+1)$$, it won't disappoint you. Regards, Leo  JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 08/19/2022, 07:05 AM Okay, Leo, I love our dialogue. Please don't feel any persecution. I think your solution, and what you've described is very valuable. But it is an asymptotic expansion; upon which Taylor data will fail. If you have an asymptotic expansion at $$z = 0$$, and you know the Taylor expansion at $$z=0$$ diverges--Then the function $$f(z)$$ can be holomorphic around values near $$z=0$$. But it cannot be holomorphic on $$|z| < \delta$$. So essentially, yes you can still have holomorphy near 0, but you can't actually have holomorphy at $$0$$. And I think this is your largest confusion. Again, I hope I'm not seeming like a dick, trying to discourage your constructions. I think they are very fucking valuable. But in this scenario, I think you need to identify that you haven't created a holomorphic function $$g(g(z)) = e^z-1$$ for $$|z| < \delta$$. You've created an asymptotic expansion (same as Gottfrieds), but you've massaged it to make it look nearly holomorphic (which is still a great accomplishment). At the risk of sounding like a broken record; $$g$$ can't be holomorphic on an $$\epsilon$$-ball centered at $$z = 0$$. BUT! It can be holomorphic everywhere around this point. Which reduces into two branches $$g^{\pm}$$ which are holomorphic on $$\mathbb{C}/(-\infty,0]$$ or $$\mathbb{C}/[0,\infty)$$. So they are holomorphic near zero--but never at zero. I don't want to seem like a dick; I'm just saying you need to reevaluate the constructions you are making in this instance. I have absolute confidence in much of your constructions. But it's okay to take a loss every once in a while. I mean, confidence and strength of conviction as a mathematician is important. But we need to understand and accept when we made a mistake. Quite frankly, I would like to better understand how you are making your asymptotic solution. Which looks fucking beautiful, and you should be proud of it. But remember, this IS NOT HOLOMORPHIC FOR $$z = 0$$. That's my only disagreement.  Leo.W Fellow   Posts: 79 Threads: 6 Joined: Apr 2021 08/19/2022, 07:57 AM (08/19/2022, 07:05 AM)JmsNxn Wrote: Okay, Leo, I love our dialogue. Please don't feel any persecution. I think your solution, and what you've described is very valuable. But it is an asymptotic expansion; upon which Taylor data will fail. If you have an asymptotic expansion at $$z = 0$$, and you know the Taylor expansion at $$z=0$$ diverges--Then the function $$f(z)$$ can be holomorphic around values near $$z=0$$. But it cannot be holomorphic on $$|z| < \delta$$. So essentially, yes you can still have holomorphy near 0, but you can't actually have holomorphy at $$0$$. And I think this is your largest confusion. Again, I hope I'm not seeming like a dick, trying to discourage your constructions. I think they are very fucking valuable. But in this scenario, I think you need to identify that you haven't created a holomorphic function $$g(g(z)) = e^z-1$$ for $$|z| < \delta$$. You've created an asymptotic expansion (same as Gottfrieds), but you've massaged it to make it look nearly holomorphic (which is still a great accomplishment). At the risk of sounding like a broken record; $$g$$ can't be holomorphic on an $$\epsilon$$-ball centered at $$z = 0$$. BUT! It can be holomorphic everywhere around this point. Which reduces into two branches $$g^{\pm}$$ which are holomorphic on $$\mathbb{C}/(-\infty,0]$$ or $$\mathbb{C}/[0,\infty)$$. So they are holomorphic near zero--but never at zero. I don't want to seem like a dick; I'm just saying you need to reevaluate the constructions you are making in this instance. I have absolute confidence in much of your constructions. But it's okay to take a loss every once in a while. I mean, confidence and strength of conviction as a mathematician is important. But we need to understand and accept when we made a mistake. Quite frankly, I would like to better understand how you are making your asymptotic solution. Which looks fucking beautiful, and you should be proud of it. But remember, this IS NOT HOLOMORPHIC FOR $$z = 0$$. That's my only disagreement. Well, thank you James. You're very nice. I think the mistake happens at the branch cut now, The 2 pieces of halfiterate coincide at 0 by their asymps, um, I kinda needa refresh my kinda old-version textbook. Regards, Leo  bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/19/2022, 06:26 PM (08/19/2022, 04:36 AM)Leo.W Wrote: Also, you can't define julia's function only by $$j(f)=jf'$$, to distinct a julia function, you'd need to define an initial value, and claim it's not multiplied by any of the form $$\theta(\alpha(z))$$ where theta is 1-periodic and alpha is the abel funct. The Julia function is considered as a formal powerseries here. The coefficients recursively obtained from the above equation are unique if we  set $$j_k=0$$  for $$k< m$$ and $$j_m=f_m$$ where  $$f(z) = z + f_m z^m + f_{m+1} z^{m+1}+ ...,\quad f_m\neq 0$$. There is no theta-ambiguity here - like with the regular iteration at a fixed point. For parabolic iteration the condition is that the iterates are asymptotically analytic at the fixed point. (08/19/2022, 04:36 AM)Leo.W Wrote: bo198214 Wrote:But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.Please lemme know what is this and why~ Its written in Milnor "Dynamics in one complex variable" And btw I can not read your quotes when you use [size=1]! Leo.W Fellow   Posts: 79 Threads: 6 Joined: Apr 2021 08/28/2022, 01:41 PM (08/19/2022, 06:26 PM)bo198214:The Julia function is considered as a formal powerseries here. The coefficients recursively obtained from the above equation are unique if we  set $$j_k="0$$  for $$k< m$$ and" $$j_m="f_m$$ where  $$f(z) = z + f_m z^m + f_{m+1} z^{m+1}+ ...,\quad f_m\neq 0$$. There is no theta-ambiguity here - like with the regular iteration at a fixed point. For parabolic iteration the condition is that the iterates are asymptotically analytic at the fixed point. Wrote: Oh! I see! now I think I'm dumb (08/19/2022, 04:36 AM)Leo.W Wrote: bo198214 Wrote:But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.Please lemme know what is this and why~ Its written in Milnor "Dynamics in one complex variable" And btw I can not read your quotes when you use [size=1]!Thanks, I read this book last year and obviously I skipped this chapter or sth... How? I just tapped the "quote" button in chrome? Can't it load in mobiles? lol I'll be careful next time Regards, Leo  bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/28/2022, 03:45 PM (08/28/2022, 01:41 PM)Leo.W Wrote: How? I just tapped the "quote" button in chrome? Can't it load in mobiles? lol I'll be careful next time Yeah, I don't know, to me also happens something similar if I copy text from somewhere into my reply, then it automatically puts some tags (I Think \[color] and another one) around it - and I wonder why my formula isn't shown correctly! Only until I press the "'view sourcecode" then I noticed the mess. So pressing "view sourcecode" is always a good thing!  bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/29/2022, 05:05 PM (08/28/2022, 01:41 PM)Leo.W Wrote: bo198214 Wrote: (08/19/2022, 04:36 AM)Leo.W Wrote: bo198214 Wrote:But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.Please lemme know what is this and why~ Its written in Milnor "Dynamics in one complex variable" Thanks, I read this book last year and obviously I skipped this chapter or sth... In this version of the book it is Theorem 7.7, Lemma 7.8 and particularly the following remark. tommy1729 Ultimate Fellow     Posts: 1,699 Threads: 373 Joined: Feb 2009 09/09/2022, 12:24 AM As for divergent summations I wanted to say something long ago but it was already said elsewhere : " repeated borel summation " @MISC {4246080,     TITLE = {Has someone seen a discussion of the (divergent) summation of $\sum\limits_{k=0}^\infty (-1)^k (k!)^2$?},     AUTHOR = {Caleb Briggs (https://math.stackexchange.com/users/709...leb-briggs)},     HOWPUBLISHED = {Mathematics Stack Exchange},     NOTE = {URL:https://math.stackexchange.com/q/4246080 (version: 2021-09-30)},     EPRINT = {https://math.stackexchange.com/q/4246080},     URL = {https://math.stackexchange.com/q/4246080} } https://math.stackexchange.com/questions...limits-k-0 *** the validity of the borel summations matches the number of solutions for fractional iterates when taking in the correct direction. nice. *** sorry if im repeating said things , i have to catch up on reading ... regards tommy1729 « Next Oldest | Next Newest »

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