11/15/2007, 08:40 AM
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]](http://tetration.itgo.com/pdf/superlog-imaginary.png)
Andrew Robbins
It depends very much on Jay's observation that slog is imaginary-periodic.
Let
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]](http://tetration.itgo.com/pdf/superlog-imaginary.png)
Andrew Robbins