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

[tommy1729]

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

I conjecture that - using mike3 branches - L and L* are the only singularities.

Also i think 0.94 + 1.12 i is a singularity but on a nearby branch.

Mike3 branches match very Well with the fake exp^[1/2].

So im intrested in the other branches ( and pics ) and also their fake analogues - although that fake should Maybe be discussed in another thread.

I think it is justitied to search the sigularities by starting from re part and then going Up or down the branches.
( like sheldon did in part )
For negative real parts there are 3 lines to start with - using mike3 branches - , not sure how to decide.

That's alot of " new " conjectures , although i had them for years actually.
Just decided to post them now.

Recently ( dec 2015 ) Sheldon started a thread about the singularity at L.
I will only talk about that there.

I wish you all a good 2016.

Regards

Tommy1729