Ya, it's a rough proof, I'm not perfectly confident in it either. But it cannot really be derived algebraically, it was to be derived like an initial value problem. It's alright to have doubts on it, I do too. But also, it appears to be unnecessary to the grand scheme. I mean, what are you to do when you run a calculator and they agree to 9 digits at 9 digits precision. Either they're deceptively close and the proof is wrong, or it just happens.
It does make sense that it would happen though, because it implies that:
\[
\alpha_{\varphi_0}\\
\]
Is a true abel function of \(x [s]_{\varphi_0} y\), it's not too crazy if you think of it like that. But it is still difficult, I agree.
Just know, I too am not satisfied with the proof, it's just a sketch of why it should happen, based on an initial value problem.
I think the key to visualizing it is tricky, which is why looking at the 1-periodic case is much more acceptable.
In the three variable case \({\bf\varphi} = (\varphi_1,\varphi_2,\varphi_3)\), we already can derive that \(\varphi_3(y-1) = \varphi_2(y)\). So to ask that it is instead \(1\)-periodic isn't much of a stretch, and there are solutions on the surface \(\Phi\) that satisfy this. The question would be, does this make a well defined operator though, or is this just solving the abel equation, without really being a true operator \((e,\infty) \times (e,\infty) \to (e,\infty)\).
So many questions, so many headaches, lmao.
It does make sense that it would happen though, because it implies that:
\[
\alpha_{\varphi_0}\\
\]
Is a true abel function of \(x [s]_{\varphi_0} y\), it's not too crazy if you think of it like that. But it is still difficult, I agree.
Just know, I too am not satisfied with the proof, it's just a sketch of why it should happen, based on an initial value problem.
I think the key to visualizing it is tricky, which is why looking at the 1-periodic case is much more acceptable.
In the three variable case \({\bf\varphi} = (\varphi_1,\varphi_2,\varphi_3)\), we already can derive that \(\varphi_3(y-1) = \varphi_2(y)\). So to ask that it is instead \(1\)-periodic isn't much of a stretch, and there are solutions on the surface \(\Phi\) that satisfy this. The question would be, does this make a well defined operator though, or is this just solving the abel equation, without really being a true operator \((e,\infty) \times (e,\infty) \to (e,\infty)\).
So many questions, so many headaches, lmao.