Extension of tetration to other branches mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 10/24/2009, 08:27 AM (This post was last modified: 10/25/2009, 10:37 PM by mike3.) Hi. I was wondering: what happens if one tries to extend the tetration to its other branches in the complex numbers, and build up the Riemann surface? Consider, for example, the regular tetration for bases $e^{-e} \lt b \lt e^{1/e}$. We get: ${}^z b = \exp^z_b(1) = \lim_{n \rightarrow \infty}\log^n_b\left(F\left(1 - \ln(b)^z\right) + \ln(b)^z \exp^n_b(1)\right)$ where $F = -\frac{W(-\ln(b))}{\ln(b)}$ is the fixed point or limit as the tower goes to infinity over the integers. (see this thread) Let us now consider what is happening with the above formula. The portion $F\left(1 - \ln(b)^z\right) + \ln(b)^z \exp^n_b(1)$ is an entire function of z for any given n. The part that begins to introduce the branch points, then, is the repeated taking of logarithms $\log_b$. Consider the removal of the first log by exponentiating both sides to base b. This causes the left to become ${}^{z+1} b$, which has no singularity at $z = -2$. So we see the outermost log creates the singularity at $z = -2$, and we can then deevelop all branches obtainable from whirling around this singularity only by adding integer multiples of the period $\omega$ of the base-b exponential, i.e. $\omega = \frac{2\pi i}{\ln(b)}$. This allows us to get $\mathrm{tet}_b_{[s_0]}(z) = \lim_{n \rightarrow \infty}\log^n_b\left(F\left(1 - \ln(b)^z\right) + \ln(b)^z \exp^n_b(1)\right) + \omega s_0,\ s_0 \in \mathbb{Z}$ for the continuation around the lowest (logarithmic) order (i.e. logarithmic) singularities. By extension of this idea, we see the next log in generates the singularities of the next higher logarithmic order (i.e. double logarithmic), and we can say $\mathrm{tet}_b_{[s_0, s_1]}(z) = \lim_{n \rightarrow \infty}\log_b\left(\log^{n-1}_b\left(F\left(1 - \ln(b)^z\right) + \ln(b)^z \exp^n_b(1)\right) + \omega s_1\right) + \omega s_0,\ s_0, s_1 \in \mathbb{Z}$ We can continue on further, and we obtain, for a general finite integer sequence $s_0, s_1, ..., s_k$, which I call a "branch code": $\mathrm{tet}_b_{[s_0, s_1, ..., s_k]}(z) = \lim_{n \rightarrow \infty}\log^n_b_{[s_0, s_1, ..., s_k]}\left(F\left(1 - \ln(b)^z\right) + \ln(b)^z \exp^n_b(1)\right)$ where we define $\log^n_b_{[s_0, s_1, ..., s_k]}(z) = \begin{cases} \log_b\left(\log_b\left(...\log_b\left(\log_b(z) + \omega s_{n-1}\right) + ...\right) + \omega s_1\right) + \omega s_0,\ \mathrm{if}\ n \le k+1 \\ \log_b\left(\log_b\left(...\log_b\left(\log_b\left(\log_b^{n-(k+1)}(z)\right) + \omega s_{k}\right) + ...\right) + \omega s_1\right) + \omega s_0,\ \mathrm{if}\ n \g k+1 \end{cases}$, and $\log_b$ are principal branches of the base-b complex logarithm. Thus the collection of all branches constructed this way for all finite sequences, I'd believe, would form the Riemann surface and the complete multivalued function for tetration to that base. Note that there are also likely branch points in the base, however continuation on that would mean going outside the regular iteration to other methods that we don't know (it's sort of like tetrating b = 0.04 to complex height) -- though I wonder if the continuum sum formula might be able to do it (seen my other thread about that?). Right now, though, I'm just focusing on the extension around the branchpoints in $z$. It may be possible to think of the sequence $[s_0, s_1, ..., s_k]$ as meaning "start at the principal branch, go and run $s_0$ times around $z = -2$, then go and run $s_1$ times around $z = -3$, and so on, finally running $s_k$ times around $z = -2 - k$". The principal branch has code $$, or $[0, 0]$, etc. One thing of note here is that on some of these other branches, points that were singularities on the principal branch may not be so on these. Consider, for example, the branch $[0, 1]$. When we make the value for $z = -2$, the next-to-outermost log has $\omega$ added to it. As its value is zero (since it represents $^{z+1} b$ as mentioned before and $^{-1} b = 0$ on the principal branch) we get the value here as actually $\log_b(\omega)$ instead of being a singularity with undefined value. $z = -3$ is still a singularity, however (it has to be, as we ran around it once to get here.). Graphs of individual branches did not seem very interesting. Many appear "flatter" than the principal branch, interestingly (except, of course, for those obtained as $\mathrm{tet}_b_{[s_0]}(z)$ with varying $s_0$). But what happens when you put them all together? Say we glued together all branches with 4-integer codes ranging from $[-4, -4, -4, -4]$ to $[4, 4, 4, 4]$ -- $9^4$ or 6561 branches, and base $b = \sqrt{2}$. What comes out? I don't have any program capable of graphing multivalued graphs, which are 3-dim (actually 4-dim, but projected to 3-dim by viewing the real and imag part separately) layered objects. And 6561 layers is a boatload of layers! Maybe it'd be easier to start with, say, $[-2, -2, 0]$ to $[2, 2, 0]$, which has $5^2$ or 25 layers, or the somewhat harder $[-1, -1, -1, -1, 0]$ to $[1, 1, 1, 1, 0]$, which has $3^4$ or 81 layers. Does anyone have anything they could use to take a crack at this? I'd love to see some pictures. There is one more property I thought worth mentioning. The formulas are given for finite sequences of integers only. By the limit process, we could define it for infinite sequences, which would generate uncountably many new values. However I am not sure whether or not these points could be truly thought of as values of tetration, because there appear to be certain theorems (see this sci.math newsgroup posting and thread I had a few months ago) that say the analytic continuation of a complex function to a multivalued function must produce only countably many values: presumably, the resulting "Riemann surfaces" from that limit procedure would contain disconnected elements (all the uncountably many sheets we just added), and so could not be interpreted as the result of analytic continuation of $\mathrm{tet}_b(z)$ in the $z$-parameter. However, even if these cannot be interpreted as proper values of the tetrational, the resulting object may still be mathematically interesting in some way... bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/24/2009, 09:54 AM (10/24/2009, 08:27 AM)mike3 Wrote: The formulas are given for finite sequences of integers only. By the limit process, we could define it for infinite sequences, which would generate uncountably many new values. However I am not sure whether or not these points could be truly thought of as values of tetration, because there appear to be certain theorems (see this sci.math newsgroup posting and thread I had a few months ago) that say the analytic continuation of a complex function to a multivalued function must produce only countably many values: ... Now that I read your article I revised my view on the number of branches, as there is an error contained in it which probably also Daniel Geisler was incorporating. I described my previous belief and the error in it here. Applied to your model of branches in the limit formula I would just guess that the limit formula does not converge if you choose infinitely man branches different from the main branch. mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 10/24/2009, 08:01 PM (10/24/2009, 09:54 AM)bo198214 Wrote: Applied to your model of branches in the limit formula I would just guess that the limit formula does not converge if you choose infinitely man branches different from the main branch. Actually it does seem to converge. The problem is that it seems to converge to the same value for every z in such cases. I.e., converging to a constant function. There are uncountably many such limit values, yet as constant functions they are "analytically incompatible" (is that a real term?) with the function (you can't analytically continue a constant function to tetration!), so they cannot be interpreted as connected Riemann sheets or Riemann branches, and thus not as values of the tet function. I'm not sure of a proof though that this is actually what happens, but some way or another, it has to fail. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/24/2009, 08:13 PM (10/24/2009, 08:01 PM)mike3 Wrote: Actually it does seem to converge. The problem is that it seems to converge to the same value for every z in such cases. I.e., converging to a constant function. Interesting! Quote: There are uncountably many such limit values, yet as constant functions they are "analytically incompatible" (is that a real term?) with the function (you can't analytically continue a constant function to tetration!) Well, each constant fixed point of b^x is a tetration! I.e. it satisfies c(z+1)=b^c(z). Does it converge to fixed points? mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 10/25/2009, 02:14 AM (This post was last modified: 10/25/2009, 03:29 AM by mike3.) (10/24/2009, 08:13 PM)bo198214 Wrote: (10/24/2009, 08:01 PM)mike3 Wrote: Actually it does seem to converge. The problem is that it seems to converge to the same value for every z in such cases. I.e., converging to a constant function. Interesting! Hmm. Sounds like it's time for a graph... I'll see if I can prepare a 2D one looking at the values of various branches on the real axis (can't do 3D with anything I've got). (10/24/2009, 08:13 PM)bo198214 Wrote: Quote: There are uncountably many such limit values, yet as constant functions they are "analytically incompatible" (is that a real term?) with the function (you can't analytically continue a constant function to tetration!) Well, each constant fixed point of b^x is a tetration! I.e. it satisfies c(z+1)=b^c(z). Does it converge to fixed points? But it's a constant function, so it cannot be interpreted as analytic continuation of the specific function $\mathrm{tet}_b(z)$ to another branch (think about the problem in "reverse": how would you analytically continue from this constant function to a non-constant one? You can't). And not all branches of $\mathrm{tet}_b(z)$ satisfy $\mathrm{tet}_b(z+1) = b^{\mathrm{tet}_b(z)}$. Consider the branch $\mathrm{tet}_b_{}(z) = \mathrm{tet}_b(z) + \omega$. We get $b^{\mathrm{tet}_b_{}(z)} = b^{\mathrm{tet}_b(z) + \omega} = b^{\mathrm{tet}_b(z)} * b^{\omega} = b^{\mathrm{tet}_b(z)} * 1 = b^{\mathrm{tet}_b(z)} = \mathrm{tet}_b(z+1) \ne \mathrm{tet}_b(z+1) + \omega = \mathrm{tet}_b_{}(z+1)$. Note that it takes us back to the principal branch. The equation $\mathrm{tet}_b(z+1) = b^{\mathrm{tet}_b(z)}$ seems to only hold for all $z$ when using the principal branch (though there may be some freedom in the choice of cut of course), if we are interpreting the symbol $\mathrm{tet}_b(z)$ as a specific single-valued branch. As you can see above, though, if we interpret it in a multivalued sense, that values on some branches of $\mathrm{tet}_b(z+1)$ equal $b^{\mathrm{tet}_b(z)}$ for values on some branches of $\mathrm{tet}_b(z)$, then it is true, as can be seen from the example I just showed. The situation is similar to that with $\log(z)$ and $\exp(z)$: $\log(\exp(z)) = z$ for some branch of log for any given $z$ but which branch that is will depend on what $z$ is. "Multivalued functions" are funny things, you know? bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/25/2009, 07:43 AM (10/25/2009, 02:14 AM)mike3 Wrote: But it's a constant function, so it cannot be interpreted as analytic continuation of the specific function $\mathrm{tet}_b(z)$ to another branch Yes you are right, its no more the regular tetration. Can you say which branch sequence you used to produce the constant function? Quote:As you can see above, though, if we interpret it in a multivalued sense, that values on some branches of $\mathrm{tet}_b(z+1)$ equal $b^{\mathrm{tet}_b(z)}$ for values on some branches of $\mathrm{tet}_b(z)$, then it is true, ... "Multivalued functions" are funny things, you know? Indeed I noticed that interpretation already in other contexts. If you have a multivalued function satisfying a certain functional equation it satisfies this equation only if you choose the suitable branches. Easiest example is log with the functional equation log(ab)=log(a)+log(b). I remember also this kind of description in [Kuzma: iterative functional equations] for regular holomorphic Abel functions. Particularly I noticed this behaviour with Dmitrii's superlog, and now you add regular tetration. mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 10/25/2009, 09:33 AM (10/25/2009, 07:43 AM)bo198214 Wrote: (10/25/2009, 02:14 AM)mike3 Wrote: But it's a constant function, so it cannot be interpreted as analytic continuation of the specific function $\mathrm{tet}_b(z)$ to another branch Yes you are right, its no more the regular tetration. Can you say which branch sequence you used to produce the constant function? I tried $[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...]$ (see the obvious pattern), $b = \sqrt{2}$ as the "infinite branch code". This seems to make it converge onto the constant function with everywhere-value ~9.21189016 + 3.20965748i. It seems any code will work. E.g. $[0, 1, 1, 1, 1, 1, 1, 1, ...]$ produces a different constant, ~9.09072986 + 3.37850978i. I suspect the Riemann surface has a interesting nested structure. Do you agree? (10/25/2009, 07:43 AM)bo198214 Wrote: Indeed I noticed that interpretation already in other contexts. If you have a multivalued function satisfying a certain functional equation it satisfies this equation only if you choose the suitable branches. Easiest example is log with the functional equation log(ab)=log(a)+log(b). I remember also this kind of description in [Kuzma: iterative functional equations] for regular holomorphic Abel functions. Particularly I noticed this behaviour with Dmitrii's superlog, and now you add regular tetration. Yeah, also consider $\exp^u(\exp^v(z)) = \exp^{u+v}(z)$ for fractional/real/complex u and v (i.e. any but natural numbers) done via, say, Kouznetsov's tetration. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/25/2009, 11:26 AM (10/25/2009, 09:33 AM)mike3 Wrote: I tried $[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...]$ (see the obvious pattern), $b = \sqrt{2}$ as the "infinite branch code". This seems to make it converge onto the constant function with everywhere-value ~9.21189016 + 3.20965748i. It seems any code will work. E.g. $[0, 1, 1, 1, 1, 1, 1, 1, ...]$ produces a different constant, ~9.09072986 + 3.37850978i. I am still not really familiar with those disconnected Riemann surfaces. Do these constants have a meaning without those limit formulas? Something path related, infinite paths, fractal paths? mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 10/25/2009, 08:47 PM (This post was last modified: 10/25/2009, 08:53 PM by mike3.) (10/25/2009, 11:26 AM)bo198214 Wrote: I am still not really familiar with those disconnected Riemann surfaces. Do these constants have a meaning without those limit formulas? Something path related, infinite paths, fractal paths? If one examines at the limit formulas, the meaning should be apparent. Each $s_k$ represents a winding number about one of the singularities. $s_0$ would represent the winding number about the singularity at $z = -2$ (and any periodicity "clone" of it), $s_1$ would represent the winding number about hte singularity at $z = -3$, $s_2$, the singularity at $z = -3$, and so on. Note that $\log_b(z) + \omega n$ is the branch of log obtained by winding $n$ times counterclockwise around the singularity at $z = 0$ from the principal branch. I'd suppose that you could think of it as "run $s_0$ times about $z = -2$, then go and run $s_1$ times around $z = -3$, then go and run $s_2$ times around $z = -4$, then go and run $s_3$ times around $z = -5$, and so on, ad infinitum and take the limiting function which all the generated branches approach". For the given infinite winding sequences, the branches approach constant functions with the limits given. I suspect it approached a constant function for every infinitely long sequence of $s_k$ (i.e. one that does not end in an infinite string of 0s). So an infinite path might be one way of imagining it, or the limit of infinitely many finite paths. In this way, meaning can be given without resorting to the limit formulas. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 10/25/2009, 09:22 PM (10/25/2009, 08:47 PM)mike3 Wrote: So an infinite path might be one way of imagining it, or the limit of infinitely many finite paths. Ya, but there are no infinite paths. Only little before I discussed that in this post. If you want to come to a branch value you need a path between two points (or a closed path). And a path between two points or a closed path can not be infinite, can it? « Next Oldest | Next Newest »

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