Arguments for the beta method not being Kneser's method

[JmsNxn]

....
Let's consider \( tet(s+c) = \lim_{n\to +\infty}f^n(s)=\ln^{\circ n}(f(s+n)) \)

t(s+n) is close to 1 for n small and going to 1 for n large.

then

f(s+n+2)= exp(t(s+n+1) * f(s+n+1))

f is never zero so log f is never log(0).

So lets investigate ln ln f.

if ln ln f = log(0) then f must be 1 exactly.

f(s+n+2)= exp(t(s+n+1) * f(s+n+1)) = 1

ln f(s+n+2) = t(s+n+1) * f(s+n+1)

since t is close to 1 , and f(s+n+1) is never zero , t(s+n+1) * f(s+n+1) is never close to 0 ! ( but rather closer to k 2 pi i for k at least 1 in absolute value.)

so ln ln(f(s+n+2)) is never log(0).

by induction f = 1,e^e,... all do not give rise to log(0).

So log singularities are not " expected ".

essential singularities are also not much expected.

So its seems close to the real positive line we get analytic. ( t(s) is close to 1 there )

regards

tommy1729

Thank you, Tommy. This is a great alternative angle. It continues to remind me there are many ways of ensuring the beta method is holomorphic. I'm still set that these are two very different ways of looking at the problem. The beta way, or the gaussian way--the additive, versus the multiplicative; even though both are technically the same through a variable change.

Edit: To Sheldon, the difference between my method and tommy's method is \( \beta(s+1) = \exp(\epsilon + \beta(s)) \) and \( \text{Tom}(s+1) = \exp((1+\epsilon)\text{Tom}(s)) \); but both are related by a change of variables--hence equivalent.