Arguments for the beta method not being Kneser's method

[tommy1729]

I could not sleep because I felt I did not clarify my previous posts enough.

The point is I showed that the first 2 logaritms do not pose a problem.

However the third and fourth logs might be problematic.

With the first 2 logs I was able to " magically " remove the logs , but with ln ln ln ln f( s + 4) this will be pretty hard.

so the 3rd and 4th log might give rise to singularities.

however this does not immediately imply that the limit has a singularity or does it ?
Lets investigate

assume
r(s) = ln^[n]  f(s+n) = ln(0)

then r2(s) = ln^[n+1] f(s+n+1) = ln^[n] ( f(s+n) t(s+n) ) 
For finite n this probably does not equal ln^[n]  f(s+n) so it does not follow automatically ...



so those two things ( 3rd and 4th log and r(s), r2(s) ) kept me awake.

The situation is still unclear and I have not even considered the "problem" of branches and periodic points etc.

also although ln(0) for a given n might not be an issue , for a larger value of n say m we might also run into log(0) so it does completely resolve things either !!!!

 
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So to think better about the branches , I need to formalize.

And that formilazation is just a proposal/conjecture because it might not be the best aka be analytic or equivalent.

f(s+1) = exp( f(s) t(s) )

then 

ln f(s+1) = f(s) t(s) = f(s + h(s))

and

exp( f( s + h(s) ) = f(s+1)

and 

f( s + h(s) + 1) = exp( f(s + h(s)))  t(s + h(s)) )

therefore

f( s + h(s) + 1)^( t(s+h(s))^{-1} ) = f(s+1).

So the 2 fundamental equations for h(s) are :

f(s) t(s) = f(s + h(s))

and 

f( s + h(s) + 1)^( t(s+h(s))^{-1} ) = f(s+1)

In a way we got rid of exp and ln here.

How many solutions h(s) do we get ??

I mentioned h(s) before but then ( i believe ) i only used the first equation. That had many solutions.

how close h(s) is to zero , what branches it implies etc is all closely related ofcourse.

Once we understand h(s) we continue ;

ln ln f(s+2) = ln ( f(s+1) t(s+1) ) = ln( f(s+1 + h(s+1)) ) = f(s+h(s+1))  t(s+h(s+1)) = f( s + h(s+1) + h(s + h(s+1)) ).

And it is clear we can continue "removing" logs with those h's.

the speed of growth of h is ofcourse very important for the convergeance and analyticity of ln^[n] f(s+n) !!

regards 

tommy1729