Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible?

[mike3]

Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.

This one seems to converge pretty poorly; much slower than ideal. If you take \( f(\text{sexp}(18\pi i))<\approx 10^{-32{ \), and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms.

For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising.

Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(z*I)) function.

Which makes one wonder: what about modifying the Cauchy integral equation so as to compute \( e^{z^2} \mathrm{tet}(z) \) instead of \( \mathrm{tet}(z) \) directly? Note that this function will still approach a limiting value at \( \pm i\infty \), in particular, 0 at both ends!

If we let \( G(z) = e^{z^2} \mathrm{tet}(z) \), then

\( G(z+1) = e^{(z+1)^2} \mathrm{tet}(z+1) = e^{z^2 + 2z + 1} \exp(\mathrm{tet}(z)) = e^{z^2 + 2z + 1} \exp(e^{-z^2} G(z)) \)
\( G(z-1) = e^{(z-1)^2} \mathrm{tet}(z-1) = e^{z^2 - 2z + 1} \log(\mathrm{tet}(z)) = e^{z^2 - 2z + 1} \log(e^{z^2} G(z)) \).

Then the Cauchy integral equation looks like

\( G_A(z) = \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 + 2ip + 1} \exp(e^{p^2} G(ip))}{1 + ip - z} dp - \frac{1}{2\pi} \int_{-A}^{A} \frac{e^{-p^2 - 2ip + 1} \log(e^{-p^2} G(ip))}{-1 + ip - z} dp \).

where \( G(z) = \lim_{A \rightarrow \infty} G_A(z) \). Note that unlike Kouznetsov's original Cauchy integral equation, there is no residual term \( \mathcal{K}(z) \) because this function approaches 0 at \( \pm i\infty \).

This is just a theory, just a guess -- I have no idea if this will work or not. I am also concerned about the possibility of ambiguity because the fixed points are not specified in this equation and so you might wind up with the alternate fixed point solution instead. I suppose one could, perhaps, avoid that by adding a residual which is the Cauchy integral of the fixed point times the Gaussian, but that integral doesn't appear to be representable in terms of any standard functions. Or perhaps by trying to restrict it by some kind of restriction involving the branches of the \( \log \) that appears there? Though the right initial guess should probably allow it to converge to the correct function, but finding a good initial guess for tetration seems tricky.

But thought I'd toss it out there anyways...