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Super-logarithm on the imaginary line
I just found a way to calculate slog on the imaginary axis!
It depends very much on Jay's observation that slog is imaginary-periodic.

Let . S is periodic with period , because . Since S is periodic, we can use Fourier series to represent it. Let , then . The Taylor series coefficients of R will then be the Fourier series coefficients of S. In terms of the super-logarithm, . This means the Fourier series coefficients of are the Taylor series coefficients of which we already know. In other words, , so:

The nice thing about this is that it seems to bypass the radius of convergence problem near z=i since its a Fourier series and not a Taylor series. Is this right?

I've included a plot with multiple approximations, which seem to converge much faster than doing analytic continuation section-by-section. The top line is the imaginary part, and the bottom line is the real part, and the "y" axis is :

PDF version

[Image: superlog-imaginary.png]

Andrew Robbins

Messages In This Thread
Super-logarithm on the imaginary line - by andydude - 11/15/2007, 08:40 AM

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