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


