slog_b(sexp_b(z)) How does it look like ?

[sheldonison]

[quote='sheldonison' pid='6865' dateline='1398462526']

Hmmm
If we take + instead of - :
.....
I changed - to + to avoid log branches.

regards

tommy1729

Because sexp(z) is analytic in the upper and lower halves of the complex plane, there is no ambiguity in log branches when looking at sexp(z), sexp(z-1), sexp(z-2), sexp(z-3), so no need to change - to +. In addition, the Taylor series gets more and more chaotic as z increases, so its easier to take slog(sexp(z-n)) for smaller z, and as long as the branches are the same, it does not change the result, except by subtracting n, so stop when the branches first differ, and then take slog(sexp(z)+2npi i). Then develop an intuitive explanation for the behavior of this function locally at z, the point in question. In particular, one can always find a path towards the fixed point of L, for sexp(z) as imag(z) increases, and then along this path, we are taking slog(L+delta+2npi i), which seems to be sufficient to show this solution does not have the property that slog(sexp(z+1))=slog(sexp(z))+1, along the path of increasing imaginary(z).

The relevant equation would be as for normal sexp(z), where slog(sexp(z))=z, \( \lambda=\ln(L) \), then as Im(z) increases, \( \text{slog}(L+\lambda\delta) = \text{slog(L+\delta)+1+O\delta \), where the trick would be coming up with a rigorous equation for the error term.

If the branches do line up, then my conjecture would be that the above approximation holds, and the solution will always be a multiple of the period as imag(z) goes to infinity. In that case, the solution would have the property that

\( \text{slog}(\text{sexp}(z))=z+\theta_n(z)+n\text{period} \), where theta is a 1-cyclic function, which goes to zero as \( \Im(z) \) goes to infinity, and theta is uniquely determined by n.
- Sheldon