eta as branchpoint of tetrational

[sheldonison]

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I wonder whether Sheldon could supply some pictures of that, with his fourier algorithm.
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I still agree with you that everything looks like there is a branchpoint, but I can not follow your particular arguments in the moment.

I've posted a few pictures of the fourier/theta(z) mapping, where I plot theta(z), at imag(z)=1i, and then again at imag(z)=0.001i. Note that \( \text{sexp}_\eta(z)=\text{cheta(z+\theta(z)+k) \), where imag(k)= 1.0472i, and real(k) is a constant to line up the two functions.


Finally, for a more detailed view at the real axis itself, I switch from plotting theta(z) to plotting the repeating contour that thata(z) is mapping in the cheta(z) function. Again, I lined the two functions up in the real direction (but not the imaginary). Here, we see two iterations of that contour function, from -1<z<0, and from 0<z<1. To get reasonable range and precision, I actually calculate the contour from -3<z<-2, which is defined by \( \text{cheta}^{-1}(z)-z \), where z varies from \( +\infty +e\pi i \) to \( -\infty + e\pi i \), which becomes the contour at the real axis for \( \text{sexp}_\eta(-3<z<-2) \). The singularities at each end of these contours would extend all the way to the right to +infinity; for this plot, I'm using 1.3E62, so they extend the contour that corresponds to 0<z<1 to around z=2. If you imagine this contour for cheta(z) infinitely repeating, then everything above the repeating contour becomes the \( \text{sexp}_\eta(z) \) function for positive imag(z). For negative imag(z)<0, we use the Schwarz reflection property.