superfunctions of eta converge towards each other

[sheldonison]

Where \( \theta(z) \) is a 1-cyclic function, which quickly decays to zero as imag(z) increases. Then, the constant "k" is 5.0552093131039000 + 1.0471975511965977*I.

Without looking at numerical values I at least want to mention, if theta is 1-cyclic and bounded (which would follow from decay towards ioo) then theta would be constant, because every bounded entire function must be constant.

PS: you have some typos in the thread title.

Yeah, if the 1-cyclic function is a sum, \( \theta(z)=\sum_{n=1}^{\infty}a_n \exp(2n\pi zi) + b_n \exp(-2n\pi zi) \), then all of the b_n terms must be zero, so it decays to zero as imag(z) increases. The theta(z) function is a Kneser mapping function, which is the sum of the a_n terms, with singularities for integer values of z. It is defined if imag(z)>0, or (if imag(z)=0 and z is not an integer). Then the sexp(z) function for imag(z)<0 is defined by a Schwarz reflection. And, just like sexp base e, the sexp_eta(z) winds up with singularities for integers <= -2, and the singularity cancels for integers>-2.

I was curious to see if anyone else had noticed this. Unfortunately, I don't know how to prove it. edit: In a prevoius post Henryk pointed out that if you have a sickle between the two conjugated fixed points, then that is a required condition. I'll have to try to understand that better ....

I have also numerically calculated the terms for theta(z), and verified that \( \text{sexp}_{\eta}(z)=\text{cheta}(z+\theta(z)+ k) \). At imag(z)=I, the upper harmonics have already decayed enough that the graph looks very sinusoidal. Here is the graph of theta(z) at imag(z)=1, from -1+i to 1+i, where the amplitude of the main harmonic has decayed to 0.00017. Closer to the real axis, as I have I previously posted, the graph has has progressively more and more high frequency terms, due to the singularities at integer values of z.
   
- Sheldon