Theory about tetrating bases from 0 to e^-e via the continuum sum

[mike3]

What do you think of my theory?

Hm, well, for me it sounds quite "constructed".
I think you should point out what criteria your super-exponential should satisfy.
I mean we have already a solution (see the predecessor thread) but somehow you dont find it satisfactory. So what kind of solution do you want? Please specify!

Well, the one that solves

\( \log_b\left(\frac{\mathrm{tet}'_b(x)}{\mathrm{tet}'_b(0) \log(b)^x}\right) = \sum_{n=0}^{x-1} \mathrm{tet}_b(n) \)

when a suitably "natural" definition of continuum sum is used, and the conditions

\( \mathrm{tet}_b(-1) = 0 \)
\( \mathrm{tet}_b(0) = 1 \)

so \( \mathrm{tet}_b(x) = \exp^x_b(1) \). If by "already have a solution" you mean the regular iteration, no, I don't consider it satisfactory, as it cannot solve the two other restrictions mentioned above, even though it should, considering the behavior of tetration elsewhere.

For the "natural definition of continuum sum", since I don't yet have a working method for defining it on a power series, one has to try something else. The idea was based on a paper about fractional sums by someone named Mueller, who mentioned the following "natural" formula one can obtain from certain sum identities:

\( \sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x) \)

provided the sums \( \sum_{n=1}^{\infty} f(n) \) and \( \sum_{n=1}^{\infty} f(n+x) \) converge, which I rewrote for the offset case:

\( \sum_{n=0}^{x-1} f(n) = f(0) + \sum_{n=1}^{\infty} f(n) - f(n + x - 1) \).

Thus I suggested a way to get the continuum sum of this \( \mathrm{tet}_b \) using the above "Mueller formula", based on the expected behavior and using that to relate it to the It turns out that Faulhaber's formula can be obtained from the above using analytic continuation with the case \( f(x) = \frac{1}{x^k} \) for \( k > 1 \). Note the Mueller formula only depends on the behavior at \( n \) and \( x + n \) for \( n = 0, 1, 2, ... \) and nowhere else. Then I use the idea that if we know \( \sum_{n=0}^{x-1} f(x) - g(x) \) and \( \sum_{n=0}^{x-1} g(x) \) at the given point, then \( \sum_{n=0}^{x-1} f(x) = \left(\sum_{n=0}^{x-1} f(x) - g(x)\right) + \sum_{n=0}^{x-1} g(x) \) even if \( \sum_{n=0}^{x-1} f(x) \) cannot be obtained directly.

Also, can you give a graph of the regular superexponential of such a base developed the repelling real fixed point? Even though it's not the tetrational we want, it might nonetheless help to get a general idea of what the behavior may be like. Helms posted a graph for \( b = 0.04 \) once, but it was a parametric plot, not a plot along the x-axis.

What plot do you want? He posted the super-exponential \( f(x)={\exp_{0.04}}^{\circ x}(0.5) \) for real \( x \).

The plot for that regular iteration, but not as a parametric, rather an x-y type plot (two graphs, I'd presume, for the real and imag part). This makes it easy to see what happens with, e.g. a unit increment starting at a point x, to see if my continuum sum idea would make any sense. The plot should range over the same x-values as the Helms' one, and the x-axis should be ticked off at every unit increment from 0 (i.e. marked at 0, 1, 2, 3, ...).

Actually, I already tried graphing this once but I'd like to see if yours looks like the one I had gotten (as I used a limit formula that is probably equivalent to the regular iteration but I don't know if it was or not).