zero's of exp^[1/2](x) ?

[sheldonison]

then sexp(-3.5) , sexp(-4.5) , ... sexp(-(2n+1)/2)

should all have a singularity because sexp(x-1) = ln(sexp(x))

right ?

yes, but I don't know where they would be in the complex plane.
sexp(-2.5)=-0.36237+iPi, and if you follow a path from -0.36237 to -0.36237+iPi, the singularity is right there (plotted the path earlier). But, for sexp(-3.5) = 1.1513+i1.6856, I'm not sure what the path would be in the complex plane. If I naively calculate slog(1.1513+i1.6856), I get 0.94439+i1.12428, which has no connection to the predicted singularity at exp^[0.5](sexp(-3.5)).
- Sheldon