10/17/2009, 02:59 PM (This post was last modified: 10/17/2009, 03:05 PM by Base-Acid Tetration.)

hmm... i think the one at -3 would locally be a DOUBLE-logarithmic singularity, and the one at -4 is a triple-logarithmic singularity, etc. (if you take the log of 0 n times, you get a log^n singularity; if you exp^n a log^n singularity you get zero.)

A logarithmic singularity is not an essential singularity!
All 3 types: removable singularity, pole and essential singularity, are isolated singularities, i.e. the function is holomorphic in vicinity except at that singularity.
This is not the case for the logarithm at 0.
0 is a branchpoint of the logarithm and of roots.

Quote:It is also known that there are essential singularities at -3, -4, etc, but exactly what kind of singularities these are is not well known.

Jay would call them doubly, trice, etc logarithmic. These all are branchpoints not isolated singularities.

10/17/2009, 09:29 PM (This post was last modified: 10/17/2009, 09:30 PM by Base-Acid Tetration.)

Not very rigorous argument:

The region around -2 is locally a logarithmic branchpoint. At -2 every branch of tetration, like those of the logarithm, falls to an infinite value. since tet(-3)= log(tet(-2)) and logarithm of infinity, whatever infinity it is, is a kind of essential infinity, so every branch of tetration has these singularities we can repeat this process to get more than one singularities for every branch of tetration, which an entire pentation must avoid, so any entire pentation is trivial, a constant function pen(z) = fixed point of tet(z).

10/23/2009, 08:01 PM (This post was last modified: 10/23/2009, 08:11 PM by andydude.)

Ok, I think I know what you were talking about now.

The primary branchpoints of pentation are going to be where you see the logarithmic singularities in the picture below. These are approximately at
\( -1.68187 \pm 1.33724 I \) which is also approximately \( {}^{\infty}e - 2 \). It is not exact, because the region between -1 and 0 is only approximately (x+1). In other words, if L is a branchpoint of the superlogarithm, then \( slog(slog(L)) \) is going to be branchpoint of pentation.

10/24/2009, 12:26 AM (This post was last modified: 10/24/2009, 12:27 AM by andydude.)

(10/23/2009, 09:27 PM)Base-Acid Tetration Wrote: Is it me, or will pentation indeed look a lot like tetration in the complex plane?

Well, if by "look like" you mean it will have 2 branchcuts (\( -1.6 \pm 1.33i \)) instead of 1 branchcut (-2), and 4 primary fixedpoints instead of 3 primary fixedpoints, then yes.

Below is a complex plot of \( \text{pexp}(z) - z \), and even though there is a bit of double-vision, it is clear that there are 4 fixedpoints near the origin.

(10/24/2009, 12:26 AM)andydude Wrote: Well, if by "look like" you mean it will have 2 branchcuts (\( -1.6 \pm 1.33i \)) instead of 1 branchcut (-2), and 4 primary fixedpoints instead of 3 primary fixedpoints, then yes.

I mean the way they both decay to a conjugate of values at large imaginary parts and positive real part.

10/24/2009, 04:12 AM (This post was last modified: 10/24/2009, 04:15 AM by andydude.)

(10/24/2009, 12:31 AM)Base-Acid Tetration Wrote: I mean the way they both decay to a conjugate of values at large imaginary parts and positive real part.

Yes, it appears that
\( \lim_{x\to\infty} e{\uparrow}^3 (i x)
= 0.7648667180537022 + 1.5298974233945777i \) (obtained with \( \text{sexp}^{\infty}(1+1.5i) \))
\( \lim_{x\to\infty} e{\uparrow}^3 (-i x)
= 0.7648667180537022 - 1.5298974233945777i \) (obtained with \( \text{sexp}^{\infty}(1-1.5i) \))
\( \lim_{x\to\infty} e{\uparrow}^3 (-x)
= -1.85035662730682 \) (obtained with \( \text{slog}^{\infty}(-1.5) \))
do you mean negative real part?