An explicit series for the tetration of a complex height Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 10/22/2016, 10:29 PM (This post was last modified: 01/14/2017, 10:01 PM by Vladimir Reshetnikov.) NOTE: The information in this post and the attachment is outdated. For more recent development, scroll down or see http://mathoverflow.net/q/259278/9550 I found an analytic function given by an explicit series, that generalizes the tetration in the sense that it reproduces its values for all positive integer heights and satisfies the same functional relation for all arguments in its domain of analyticity. Numerically it seems to match Kneser's solution, a PARI/GP program for which was posted on this forum. Any questions and critique are welcome. A short note explaining my solution is attached and is also available at https://tinyurl.com/tetration Attached Files   Tetration.pdf (Size: 36 KB / Downloads: 694) sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 10/23/2016, 02:50 PM (This post was last modified: 10/23/2016, 04:43 PM by sheldonison.) (10/22/2016, 10:29 PM)Vladimir Reshetnikov Wrote: I found an analytic function given by an explicit series, that generalizes the tetration in the sense that it reproduces its values for all positive integer heights and satisfies the same functional relation for all arguments in its domain of analyticity. Numerically it seems to match Kneser's solution, a PARI/GP program for which was posted on this forum. Any questions and critique are welcome. A short note explaining my solution is attached and is also available at https://tinyurl.com/tetration Strictly speaking, Kneser's solution only applies to real bases greater than exp(1/e) and my inclusion of the Schröder based solution for real bases 0$ and $\rho,\tau \in \mathbb{R}^+$ with the additional restriction that $\tau < \pi/2$ This follows because if we had a second function it would be subject to all of the above steps and in the final expression (1) defining it it would necessarily equal the original function. Your solution $F$ is first of all periodic with imaginary period. This follows because $\binom{z}{n}_q$ is periodic in $z$, and your expression is an infinite sum of these terms. So it is bounded as $\Im(z)$ grows.  Secondly your solution tends to $-W(\log(a))/\log(a)$ as the real part of z grows, because $\binom{z}{n}_q \to C(q)$ as $\Re(z) \to \infty$, leaving only a constant in your series. Therefore your solution is bounded when $\Re(z) > 0$, and secondly interpolates the natural iterates, $F(n) = (^na)$. Therefore your function is indeed the standard tetration function that I laid out above. Quite frankly I am amazed by this equation you've put in front of me. In case you didn't know, I think this should in fact work on a much larger scale and produces a new way of iterating functions.  If $\phi(\xi):G \to G$ and $\phi(\xi_0) = \xi_0$ and $0 < q =\phi'(\xi_0) < 1$, then your expression works to find $\phi^{\circ z}(\xi):I \to I$ for $\Re(z) > 0$ where $\phi^{\circ z_0}(\phi^{\circ z_1}(\xi)) = \phi^{\circ z_0 + z_1}(\xi)$ , and $I$ is the immediate basin about $\xi_0$ (largest connected domain about $\xi_0$ where $\lim_{n\to\infty}\phi^{\circ n}(\xi) = \xi_0$). IF $F(z,\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q \binom{n}{m}_q q^{\binom{n}{2}} \phi^{\circ m}(\xi)$ and $F(n,\xi) = \phi^{\circ n}(\xi)$ then $F(z,\xi) = \phi^{\circ z}(\xi)$ This follows exactly how your tetration function is shown to be tetration! The big question, is how do you show $F(n,\xi) = \phi^{\circ n}(\xi)$? What a great formula! Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 01/12/2017, 11:55 PM Thanks for your elaboration. I think it is supposed to be $q^{\binom {n-m} 2}$ rather than $q^{\binom n 2}$ in the first formula (the definition of $F(z)$). Here is a link to my Mathematica program that uses the proposed formula to numerically calculate tetration values: http://goo.gl/03L8dV JmsNxn Ultimate Fellow Posts: 1,065 Threads: 121 Joined: Dec 2010 01/13/2017, 12:46 AM (01/12/2017, 11:55 PM)Vladimir Reshetnikov Wrote: Thanks for your elaboration. I think it is supposed to be $q^{\binom {n-m} 2}$ rather than $q^{\binom n 2}$ in the first formula (the definition of $F(z)$). Here is a link to my Mathematica program that uses the proposed formula to numerically calculate tetration values: http://goo.gl/03L8dV Yes, you're right. Just a typo . Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 01/13/2017, 01:12 AM Some follow-up conjectures: http://mathoverflow.net/q/259467/9550 JmsNxn Ultimate Fellow Posts: 1,065 Threads: 121 Joined: Dec 2010 01/13/2017, 07:47 PM I was wondering, could you show to me exactly how you're proving that your series equals tetration on the naturals? This is the only thing I don't quite understand. I'm mostly curious because I am pretty certain that this should work on more functions than just tetration. For example, this is just a conjecture, but I think the following should hold $\phi^{\circ z}(\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q\binom{n}{m}_qq^{\binom{n-m}{2}} \phi^{\circ m}(\xi)$ when $\phi(\xi_0) = \xi_0$ and $\phi'(\xi_0) = q$ and $\xi$ is in a neighborhood of $\xi_0$. This would be a great series representation of the fractional iterate of arbitrary $\phi$, fast converging and everything.  TY Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 01/13/2017, 08:30 PM (01/13/2017, 07:47 PM)JmsNxn Wrote: I was wondering, could you show to me exactly how you're proving that your series equals tetration on the naturals? This is the only thing I don't quite understand. Because the properties of q-binomial coefficients, for natural arguments only a finite number of terms in the series are non-zero, i.e. its partial sums eventually stabilize (so the convergence is trivial). Then it is possible to prove by induction that these sums reproduce discrete sample values from which the series is built. Basically, it means that the direct q-binomial transform of a discrete sequence can be undone by the reverse q-binomial transform. The value of q does not matter here, it is only significant for convergence of the series at non-integer arguments. I can write it in more details later, if you want. « Next Oldest | Next Newest »

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