(06/18/2014, 10:41 PM)sheldonison Wrote: The flaw in your proof is that exp^[k] may not be analytic, if k is not an integer, even if sexp(z) is analytic. If k is an integer, exp^[k] is well defined. But you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". A simple counter example to your proof is f(z), which has f'(n)=0 and f''(n)=0 for all integers>-2.
\( f(z) = \text{sexp}(z - \frac{sin(2\pi z)}{2\pi})\;\;\; f(z+1)=\exp(f(z)) \)
The reason why is because \( f^{-1}(z) \) has a cube root branch singularity for n>=0, at \( z=\exp^{[n]}(0) \). This is relevant since the exp^[k](z) function used implicitly assumes \( f^{-1}(z) \) is analytic.
\( \exp^{[k]}(z) = f(f^{-1}(z)+k) \)
Also, the f(z) function is the "seed" value used to generate tet_alt(z) from the secondary fixed point, and is a rough approximation for tet_alt, and has the same tet'(n)=0 and tet''(n)=0 for integers>-2.
Aha but then its a partial misunderstanding.
You see , I indeed assumed \( f^{-1} \) to be analytic.
I believe that Inverse_Kneser is analytic.
SO I guess the next step is
To do : prove that if an sexp function has no singularities of the real line , then its inverse (slog) is analytic.
But I guess that if Kneser does have derivatives equal to 0 we are back at square 1.
--- EDIT ---
OK I see sheldon already said that in post 19. Sorry.
--- EDIT ---
Hmm
Its getting complicated.
Need to think more.
regards
tommy1729
