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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 \(f(z)=cze^{\frac{z1}{c}}\big(\frac{z1}{c}\big)^{a1}\Gamma(1a,\frac{z1}{c})\), 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
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(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 \(f(z)=cze^{\frac{z1}{c}}\big(\frac{z1}{c}\big)^{a1}\Gamma(1a,\frac{z1}{c})\), 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 Sure 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 \(e^z1\) is different from that by the petals of \(\log(z+1)\), it won't disappoint you.
Regards, Leo
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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 \(z=0\) divergesThen 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 \(g(g(z)) = e^z1\) 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 zerobut 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.
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(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 \(z=0\) divergesThen 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 \(g(g(z)) = e^z1\) 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 zerobut 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 oldversion textbook.
Regards, Leo
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(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 1periodic 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 thetaambiguity 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 AbelFunctions/FatouCoordinates 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]!
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(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 thetaambiguity 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 AbelFunctions/FatouCoordinates 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
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(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!
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(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 AbelFunctions/FatouCoordinates 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.
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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...lebbriggs)},
HOWPUBLISHED = {Mathematics Stack Exchange},
NOTE = {URL:https://math.stackexchange.com/q/4246080 (version: 20210930)},
EPRINT = {https://math.stackexchange.com/q/4246080},
URL = {https://math.stackexchange.com/q/4246080}
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https://math.stackexchange.com/questions...limitsk0
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the validity of the borel summations matches the number of solutions for fractional iterates when taking in the correct direction.
nice.
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sorry if im repeating said things , i have to catch up on reading ...
regards
tommy1729
