closed form for regular superfunction expressed as a periodic function

[sheldonison]

Hi Sheldon -

I recognize your coefficients....

Gottfried, thanks for your reply. I'm trying to make sense of the coefficients, and would probably need to program the matrix into pari-gp to verify it.

Earlier, Gottfried wrote:

Hi Sheldon -

just to allow me to follow (think I can't involve much) - I don't have a clue from where this is coming, what, for instance, is L at all? I think you've explained it elsewhere before but don't see it at the moment... Would you mind to reexplain in short or to provide the link?

Gottfried

L is the fixed point of base(e). If you iterate the natural logarithm function hundreds of times (say starting with z=0.5), you get a very good approximation of L. Then the equations describe the regular superfunction for base e, which is complex valued at the real axis, but is analytic and entire.

The rest of what I did basically, it comes down to expressing the regular superfunction as an analytic Fourier series, where all of the terms of the series decay to zero at i*infinity. Such a series is guaranteed to be analytic.

Here is an example of a generalized view of such a Fourier series, not related to the superfunction, with a period of 2Pi.
\( f(z) = \sum_{n=0}^{\infty}a_ne^{i*nz} \)
A full Fourier series would also have the negative coefficients as well, and is often not analytic.
\( f(z) = \sum_{n=-\infty}^{\infty}a_ne^{i*nz} \)
Both of these can be wrapped around the unit circle, with the substitution \( y=e^{iz} \). In the analytic case, we have an analytic Taylor series. \( f(z) = \sum_{n=0}^{\infty}a_ny^n \). In the general case, with terms from -infinty to +infinty, we have a Laurent series, with singularities inside the unit circle, and an annular ring of convergence. For the full Fourier series, often, the Laurent series only converges on the edge of the unit circle. Hope that helps.
- Sheldon