12/26/2009, 01:44 AM
(This post was last modified: 12/26/2009, 02:11 AM by sheldonison.)
(12/25/2009, 08:51 PM)mike3 Wrote: I think I might have an idea about what's going on here. This shows the result of graphing \( \mathrm{tet}(z) \) in the complex plane. The x-scale (real part) runs from 2 to 10 and the y-scale (imag part) runs from -8i to 8i. This function was obtained via the Cauchy integral (Kouznetsov's construction):Great plots, thanks. Do you use "mathematica"? The only tools I'm using are excel, and "perl".
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As one can easily see, they can't even be translated to fit each other, not even in the asymptotic. Though they seem like they're asymptotic on the real line, on the complex plane, they're very different, which may be why the constructed "base change tetrational" is not analytic (Taylor series has 0 convergence radius). Smooth functions, analytic nowhere, are well known, and it is also well known that in Real analysis, a sequence of real-analytic functions can converge pointwise to a non-analytic function, just look at Fourier series, where the convergent may not even be continuous!
I like the "eta" plot. I've plotted some of the n*e*pi contour lines of eta in the complex plane, which all iterate to i=0 contour lines, and which seem to be the defining features of the eta graph, along with the real=e contour lines. I'll probably post some time.
The two functions, sexp_eta, and sexp_e, seem like they should be assymptotic on real number line, but its a ringing pattern. And, as you point out, in the complex plane, they're completely different functions. Also, I now realize that the \( \theta(x) \) to which f(x) converges to is probably infinitly differentiable, but not be analytic. If you shift sexp_eta so that its super-exponential growth lines up with sexp_e, by taking sexp_eta(x+0.584) and sexp_e(x), the two graphs will intersect each other an infinite number of times, as they both super-exponentially climb to infinity, each on a slightly different pattern.
- Shel
