Trying to get Kneser from beta; the modular argument
#3
It's important to remember; the correct manner of computing \(\tau\) is a small little trick:

\(
\beta(s) = \exp^{\circ n}(\beta(s-n) + \tau^{n}(s-n))\\
\)

This is the exact implicit manner that we'd be required to do to get Kneser. It's largely slow and, not very efficient. But mathematically is what we need to remember. Where we'll start to see \(\tau^{n}(s-n) \approx L\) and \(\beta(s-n) \approx 0\).


Messages In This Thread
RE: Trying to get Kneser from beta; the modular argument - by JmsNxn - 03/29/2022, 06:34 AM

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