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(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

- Sheldon

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(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.

Posts: 1,370

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Im reminded of my old tommy's tetration constant.

TT constant for short or now maybe Tommy-Sheldon base constant.

Anyway the issue is that to make the periodicity stronger we make sure the fixpoint and period work together.

Thus im(fix) = im(period)

Exp(a z) = z

Im(z) = 2 pi / a

Then slog base exp(a) has all its singularities lined up with the desired asymptotic period.

Isn't that Nice ?

I think i talked about closed forms for a and exp(a) but I do not remember the results if any. I think it was not here.

A closed form would be appreciated.

Lambert-W ? Contour integral representation ? Nice taylor series ? Tommy-kouznetsov expansion ??

Btw I wonder how many constants are defined in this forum/tetrarion.

Regards

Tommy1729

Posts: 639

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Joined: Oct 2008

(05/14/2015, 05:58 PM)tommy1729 Wrote: (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.

Everywhere until you hit the singularity. The singularity is at L,L*. To the left of those singularities, slog(z) is both analytic and

exactly 2pi i periodic. To the right, slog is no longer 2pi i periodic. A similar 2pi i periodic would be

; it is exactly 2pi i periodic in the left half of the complex plane, but not the right half.

- Sheldon

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That is not consistent with the many singularities ...

Unless you take your slog branches different ?!

If you take your branches perpendicular to the real axis instead of parallel.

However , are we really Free to choose our branch cuts ?

Afterall we need the equality exp^[1/2](z) = sexp(slog(z)+1/2) to hold.

Regards

Tommy1729

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Joined: Feb 2009

Are you saying half-exp is periodic on the left ?

Hmm in that case there is hope for the modified 3rd question.

Need to think.

What about this Tommy-Sheldon constant ?

Regards

Tommy1729

Ps what you say about sexp is correct.

PPS : why dont we have a singularity at ln(L+2 pi i) ?

Does this imply that ln(L+2 pi i) is not in the range of sexp ? I think so !