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 Bases for which Tetration is well-defined andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/09/2007, 12:27 AM There are essentially 3 overlapping domains over which tetration is well-defined. These domains are of the form $B \times A$ where B is the set of bases and A is the set of heights over which tetration is defined. But for the sake of simplicity, we will only be talking about B since it is more interesting. The integer domain (complex bases, integer height) The regular domain (regular bases, complex height) The natural domain (natural bases, complex height) These three regions are obviously overlapping, since the set of regular bases form a subset of the complex plane, and the integers form a subset of the complex plane as well, so the first two are overlapping. The regular bases include $[e^{-e}, e^{1/e}]$ and the natural bases include $(1, \infty)$, so the intersection of these is $(1, e^{1/e}]$, so the second two have some overlap as well. If we were to take all of these domains and being them together, the union of all of these is: $\begin{tabular}{rl} B_R & = \{ b \text{ where } |\log({}^{\infty}b)| \le 1 \} \\ B_N & \approx \{ b \text{ where } \text{Re}(b) - 2|\text{Im}(b)| > 1 \} \\ B \times A & = \{(b, a) \text{ where } \left\{ \begin{tabular}{ll} a \in \bb{C} & \text{if } b \in (B_R \cup B_N) \\ a \in \bb{Z} & \text{otherwise } \end{tabular}\right\} \text{ for all } b \in \bb{C} \} \end{tabular}$ In the picture below, the horizontal is the real axis, the vertical is the imaginary axis, red means the regular region, and blue means the natural region. These are the set of all b such that ${}^{a}b$ is well-defined for all complex a (it is possible to define tetration outside this region with either method, but who knows what discontinuities might happen): The region in the complex plane over which regular tetration is well-defined is the region in which $|\exp_b'(h)| \le 1$ (where h is the fixed point) if it is less than 1 it is known as the (open) hyperbolic region, and if it is equal to 1 it is known as the parabolic region, and union of these two regions is known as the (closed) regular region. The region in the complex plane over which natural tetration is well-defined is unknown. However, I have been able to approximate the region experimentally. This region is given above as $B_N$. This region is conservative, and the region over which natural tetration actually works may be larger than this (for example the 2 might be closer to 1). The unions of these two regions covers all of the positive real line except for $(0, e^{-e})$ but before we get disappointed, I have a comment. Outside the regular domain, regular tetration is still defined and calculable, but gives complex values for the real line. It is this that makes us avoid it in preference to natural tetration on the real line above eta. Since regular tetration gives complex values for non-integer heights where the base is in $[e^{-e}, 1)$, it matters less that regular tetration gives complex values in $(0, e^{-e})$ since it makes all bases in (0, 1) complex valued. So this brings up an interesting point. There are essentially three kinds of domains we should be considering for each method of calculating non-integer tetration: Over what domain is the method be well-defined? (objective, the red/blue regions) Over what domain can the method be defined? (objective, most of the complex plane) Over what domain does the method give tetration? (subjective) I believe that these three sets are independant of each other. For example with regular tetration, the domain over which it gives tetration is the set $B_R$ defined above, but it can be defined over all complex bases. Using the other set as an example, the domain over which the natural method is defined is for all complex b subject to the conditions: The Abel matrix is invertible. The natural Abel function exists. (coefficient convergence) The natural Abel function is analytic. (series convergence) Another thing to note about these sets is the following. Where method X can be defined is completely determined, and can be as large as the complex plane, subject to the conditions of the method. The domain over which method X corresponds to tetration is entirely up to us, as we are in the process of defining tetration. Maybe since we have a method that works well for complex numbers (regular), and a method that works well for real numbers (natural), we could use the natural method for bases near the positive real axis, and use the regular method for all other complex bases, despite the fact that the bases are outside the "regular domain". The nice thing about this is that although it might be discontinuous between the "regular" and "natural" methods, it defines tetration for all complex bases, so tetration can become a piecewise-holomorphic function: ${}^{a}b\text{ : } \bb{C} \ \times\ \bb{C} \rightarrow \bb{C}$ with 2 pieces as opposed to an infinite number of pieces in the linear approximation of tetration. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/10/2007, 04:31 AM andydude Wrote:$B_N \approx \{ b \text{ where } \text{Re}(b) - 2|\text{Im}(b)| > 1 \}$ PS. This region might be as large as: $B_N \approx \{ b \text{ where } \text{Re}(b) > 1 \}$ I'm still investigating the stability of my solution with these bases. Andrew Robbins GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/31/2008, 03:21 PM Hi Andrew! Concerning: andydude Wrote:There are essentially 3 overlapping domains over which tetration is well-defined. These domains are of the form $B \times A$ where B is the set of bases and A is the set of heights over which tetration is defined. But for the sake of simplicity, we will only be talking about B since it is more interesting. The integer domain (complex bases, integer height) The regular domain (regular bases, complex height) The natural domain (natural bases, complex height) ... In the picture below, the horizontal is the real axis, the vertical is the imaginary axis, red means the regular region, and blue means the natural region. These are the set of all b such that ${}^{a}b$ is well-defined for all complex a (it is possible to define tetration outside this region with either method, but who knows what discontinuities might happen): Could you please, for my parsonal use, be kind enough to better precise what do you mean by the regular and natural bases. Thanks. GFR andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 03/30/2008, 07:09 AM (This post was last modified: 03/30/2008, 03:58 PM by andydude.) These domains are developed on two separate lines of thought, the regular one and the natural one. I'll start with the regular one. When considering geometric series, the general rule is: $\sum_{k=0}^{n-1} a r^k = \begin{cases} a n & \text{if } r = 1 \\ a \frac{1 - r^{n}}{1 - r} & \text{otherwise} \end{cases}$ while the limit as n goes to infinity is: $\sum_{k=0}^{\infty} a r^k = \begin{cases} \frac{a}{1 - r} & \text{if } |r| < 1 \\ \infty & \text{otherwise} \end{cases}$ In Szekeres' and Henryk's discussion of the regular method, the Abel function is defined in terms of the Schroeder function (which I have tried to codify here for future reference) as follows: $ \sigma(x) = \lim_{n\rightarrow\infty} \frac{\exp^n(x)}{\exp'(x_0)^n}$ where many series within the regular approach can be defined in terms of the Schroeder function, with inverses, logarithms, and so on, we can get the Abel function, the general iterate, etc. The line of thought that prompted this thread was that the Schroeder function is defined as a limiting case of an iterated function. While one could define the Schroeder function in terms of Geisler's coefficients, causing no limit of domain, the problem of defining the coefficients rests with Geisler's method which not many completely understand. I think there are more people who understand the Schroeder function than Geisler's method, so it makes more sense to defined things in terms of it. This is where the problem comes in. Geisler's coefficients are finite geometric series. The Schroder function has coefficients which are also finite geometric series, as can be seen in this post I made regarding them. However, the process of defining the Schroeder function in terms of a limit requires that we use the $n\rightarrow\infty$ version of the geometric series formula, which means that (|r|<1) instead of (r/=1) for the finite case. So while the coefficients of the Schroeder function may be finite series (r/=1), the process of finding/defining these coefficients involves a limit to infinity, which would essentially be an infinite series (|r|<1). Somewhere in here is my reason for limiting the domain of regular tetration to bases which lie in the convergence region of infinitely iterated exponentials. Ok, now for the natural one. The 3 major problems I had with natural tetration, is with proving that the coefficients are solvable, the coefficients converge, and that the series formed from these coefficients converges. One thing that I noticed recently that I'm still very interested in, but have not verified completely is as follows. Using the notation $ \begin{tabular}{rl} \text{slog}_b(x) & = \sum_{k=0}^{\infty} s_k(b) x^k \\ \text{slog}_b(x)_n & = \sum_{k=0}^{n} s_{nk}(b) x^k \end{tabular}$ I noticed that for all bases for which the coefficients converged, that $|s_1(b)| > |s_k(b)|^{1/k}$ for all k. I'm still researching this, so don't take my word for this yet... But using that observation, I thought perhaps I could turn it around, and say that the bases for which the natural super-logarithm are well defined, are exactly those bases which satisfy $|s_1(b)| > |s_k(b)|^{1/k}$. This makes the proof almost trivial! Since all you have to do is divide the super-logarithm by $|s_1(b)|$, and then the root-test of the coefficients would be bounded by a constant. As for the "triangle" above, I have no justification for why it would be a triangle, but my tests for finding where the root-formula hosts true have indicated that those are the bases for which it is most obvious that the root-formula holds true. So the question still remains, what are the bases for which the natural tetration/super-logarithm is well-defined? Andrew Robbins PS. I'm sorry for the lack of rigor in the previous posts, these ideas are somewhat vague, but I have tried to provide as much rigor as I can considering these are open problems... andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 03/30/2008, 04:01 PM To clarify even more, I think that the domain over which the super-logarithm coefficients are solvable is much bigger, specifically $|b|>1$, but since for many of these bases, the coefficients failed to converge, I had to put more thought into exactly where they converge. That is how I found the root-test above. Andrew Robbins bo198214 Administrator Posts: 1,384 Threads: 90 Joined: Aug 2007 06/24/2008, 12:41 PM Perhaps the question does not fit completely here, but I remember that you somewhere mentioned (can not find it anymore) that there is a (complex) singularity for the nslog that limits the convergence radius to some value. Can you just give this explanation/pictures again or the reference to your post? bo198214 Administrator Posts: 1,384 Threads: 90 Joined: Aug 2007 06/28/2008, 05:17 PM bo198214 Wrote:Perhaps the question does not fit completely here, but I remember that you somewhere mentioned (can not find it anymore) that there is a (complex) singularity for the nslog that limits the convergence radius to some value. Can you just give this explanation/pictures again or the reference to your post? Ah now I found the relevant post « Next Oldest | Next Newest »

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