Generalized recursive operators

[andydude]

May I ask if anyone has an idea how much this result depends on the actually chosen tetration extension to real numbers? I mean the b[n]-k=-k+1 for 0\( \le \)k\( \le \)n-3 does not depend on the extension.

Surely. Just as \( b[4](-\infty) \) is the fixed point obtained by iterating \( \log_b(x) \), so is \( b[5](-\infty) \) the fixed point obtained by iterating \( \text{slog}_b(x) \) which also corresponds to the fixed point of \( {}^{x}b \). This works well when the fixed point is an attracting fixed point, but poorly for a repelling fixed point.

For \( 1 < b < \eta = e^{1/e} \) tetration has 2 fixed points, and for \( b > B \) (does \( B = \eta \)?) tetration has only the lower fixed point. Since the lower fixed point of tetration falls between two integers is does depend on which extension is used.

To summarize, one thing we can know for sure regardless of which extension is used, is that even hyper-exponentials follow: \( b [N] (-Y) = (-\infty) \) and odd hyper-exponentials follow: \( b [N] (-\infty) = (-X) \). All odd hyper-exponentials should be real-valued over all reals. All even hyper-logarithms should be real-valued over all reals. All hyper-exponentials above 3 map [-1,0] -> [0,1] and all hyper-logarithms above 3 map [0,1] -> [-1,0]. Another common property is that the range of odd hyper-exponentials is bounded below, just as the domain of even hyper-logarithms is bounded below.

Since these findings are all using negative hyper-exponents, then they are essentially specifying the number of times to iterate the appropriate hyper-(N-1)-logarithm. All odd hyper-exponentials are real-valued over all reals BECAUSE the hyper-(N-1)-logarithm is also real-valued over all reals. One conclusion we can draw from the statements above is that the domain of even hyper-exponentials is bounded below, because the domain of odd hyper-(N-1)-logarithms is bounded below. Accordingly, the range of odd hyper-exponentials is bounded below, because the range of even hyper-(N-1)-logarithms is bounded below.

I hope that made sense. I will try and make this more formal and more clear in a further post.

Andrew Robbins