Searching for an asymptotic to exp[0.5]

[tommy1729]

....
f(x) is the half-iterate from the sinh method.

I assume since sinh is an odd function, that f(x), the asymptotic half iterate, would also be an odd function, though this is not required. As I remember, the closest singularity to the origin for sinh^{0.5} is on the imaginary axis.

For f(x) asymptotic to exp^{0.5}, the branch cut is on the negative real axis. For the asymptotic to sinh^{0.5}, one possible branch cut is on the imaginary axis, in both directions, which would lead to an odd function, with even Taylor series coefficients=0. Is this what you had in mind?

Just as you use kneser in post 9 , I use the 2sinh here.
So just for numerical reasons.
Im still on the real line.

It is convenient since it satisfies ln(f(exp(x))) = f(x) what simplifies the equations.

regards

tommy1729