05/14/2015, 05:58 PM

(05/14/2015, 03:13 PM)sheldonison Wrote:(05/14/2015, 02:28 PM)tommy1729 Wrote: ... Unless the functional equations no longer hold there.Consider that slog(0)=-1, slog(1)=0

given the 2pi i periodicity, then

slog(2pi i)=-1; slog(1+2pi i)=0

slog(2npi i)=-1; slog(1+2npi i)=0 for any value of n

sexp is the inverse of slog. Therefore, somewhere on the sexp(z) Riemann surface, as you circle around the singularity at -2:

sexp(-1)=0; sexp(0)=1

sexp(-1)=2pi i; sexp(0)=1+2pi i

sexp(-1)=2npi i; sexp(0)=1+2npi i for any value of n

Well approximately yes.

Analytic continuation forces perfect periodicity to extend everywhere.

You might like my next post.

Regards

Tommy1729

ITS hard to convince a smart person but iTS near impossible to convince an idiot.