(08/27/2009, 05:36 PM)bo198214 Wrote:I haven't read the original (I think I've maybe skimmed over it, but my German isn't fluent). I'm still going through your description, and while I seem to have a rudimentary understanding of each step in the construction, I apparently don't understand them well enough to string the steps together mentally and have one of those "Ah-ha!" moments. I'm close, but I need to read through it again and work some of the steps out myself to get a better feel for them.(08/27/2009, 04:38 PM)jaydfox Wrote: As for Kneser's construction, I gave an initial look and was quickly overwhelmed; it will take me some time to properly decipher it, so I can't really comment on it yet.Did you read the original article? You know I summarized his article here. Kneser exactly showed the construction of a real-valued superlogarithm from the regular one. The regular slog maps the upper halfplane to some infinite region D (because it has singularities on the real axis at \( \exp^{[n]}(0) \)). Now from the Riemann mapping theorem for any two regions (except the whole plane) there is a biholomorphic mapping between them. Kneser uses this biholomorphic mapping \( \phi \) to map D back to the upper halfplane. And shows that \( \phi(z+1)=\phi(z)+1 \), hence also \( \operatorname{kslog}(z)=\phi(\operatorname{rslog}(z)) \) satisfies the Abel equation \( \operatorname{kslog}(\exp(z))=\operatorname{kslog}(z)+1 \).
kslog maps G (the region bounded by L1 and exp(L1)) biholomorphically to an imaginary unbounded region. I showed that this condition (about which we would agree that the islog also satisfies it) is a uniqueness criterion here. So why am I arguing against its uniqueness? Because I finally accept statements only with proof!
But it is interesting, as I'm convinced of the uniqueness of the islog, and given that Kneser's construction seems to fit as well, I'm assuming they are numerically equivalent. I'm assuming it maintains the approximation of the regular slog near the fixed point? This is key: if it does, then it has the properties of the islog, and by the uniqueness criterion it should match the islog. And if so, I think I rather prefer this approach, as to me it's more intuitive (ironically!) than the intuitive slog (or rather, than the matrix solution method, corresponding to solving an infinite system of linear equations). If, that is, I'm understanding it correctly, which as I admit, I might not be.
You mentioned somewhere having a means to compute values for Kneser's slog: perhaps we should see how well they match up with the islog? That, or derive a power series expansion for Kneser's solution and see how it matches the islog.
And you mention the uniquess criterion that you outlined in that paper: are you saying then that you don't consider this criterion proven?
~ Jay Daniel Fox

