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.
I also plotted two branch systems:
The function being plotted is actually:
\( f(z) = \begin{cases}
\text{slog}(f(z + 1)) & \text{if } Re(z) < 0 \\
{}^{z}e & \text{if } 0 \le Re(z) \le 1 \\
\text{sexp}(f(z - 1)) & \text{if } Re(z) > 1
\end{cases} \)
Andrew Robbins
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.
I also plotted two branch systems:
- A, from branchpoints to imaginary infinities
- D, from branchpoints to negative infinity
The function being plotted is actually:
\( f(z) = \begin{cases}
\text{slog}(f(z + 1)) & \text{if } Re(z) < 0 \\
{}^{z}e & \text{if } 0 \le Re(z) \le 1 \\
\text{sexp}(f(z - 1)) & \text{if } Re(z) > 1
\end{cases} \)
Andrew Robbins