An explicit series for the tetration of a complex height JmsNxn Ultimate Fellow Posts: 1,178 Threads: 123 Joined: Dec 2010 01/14/2017, 12:36 AM (This post was last modified: 01/14/2017, 01:06 AM by JmsNxn.) (01/13/2017, 08:30 PM)Vladimir Reshetnikov Wrote: (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.Ohh! So its the convergence of the infinite sum that forces the specific value of q. I was curious as to why it wouldn't converge for other q. I was thinking there was something more complicated, but inverting the q-binomial expansion is very straightforward. Now it makes a lot of sense. Secondly, do you have any thoughts on how to show convergence. This seems to be the only thing blocking the path. I tried using the same method I used to show convergence of the newton series, but t doesn't apply to the more general q case. No where in my attempts was I using that q is the multiplier. which causes me great discomfort. I really like this expansion, so I'm stuck figuring. Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 01/14/2017, 02:13 AM (This post was last modified: 01/14/2017, 02:21 AM by Vladimir Reshetnikov.) (01/14/2017, 12:36 AM)JmsNxn Wrote: Secondly, do you have any thoughts on how to show convergence. This seems to be the only thing blocking the path. Unfortunately, I do not yet have a proof of convergence. But the intuition that lead me to this series is as follows. The difference between a tetration value of a finite integer height $^n a$ and its limiting value (fixpoint) $L={^\infty a}$ decays exponentially, where the base of the exponent (I named it q, but it is commonly denoted by λ) is the logarithm of the limiting value. Now we want to find an expression for tetration ${^z a}$ of a complex height z. Exponential decay is not approximated well by polynomials, so we need to switch from z to a new variable $w=q^z$. After this, the values of the discrete tetration all lie almost on a straight line. It seemed plausible that they could be interpolated by Lagrange interpolating polynomials without causing Runge's phenomenon (erratic oscillations between sample points). Because with the new variable w sample points are not equally spaced, but rather form a geometric progression, the products in numerators and denominators of terms in the Lagrange interpolating polynomial turn into q-Pochhammer symbols and then can be combined into q-binomial coefficients (so that's the moment when q comes into play). The series I proposed is just the limit of the Lagrange interpolating polynomial when number of sample points tends to infinity, and its partial sums are the Lagrange interpolating polynomials built on a finite number of points. By construction, the series converges and exactly reproduces sample values at integer points. Convergence at non-integer points (that I can confirm numerically, but cannot prove) is basically equivalent to not having Runge's phenomenon during polynomial interpolation. Obeying the functional equation for tetration at all points, including non-integer, does not immediately follow from the construction, but was initially just a nice conjecture supported by numerical computations. JmsNxn Ultimate Fellow Posts: 1,178 Threads: 123 Joined: Dec 2010 01/14/2017, 02:48 AM Oh, so it's because binomials of $q^z$ can approximate $\phi^{\circ z}(\xi)$. That's really clever. I like that. I can't imagine how you prove this using the method I used to prove the newton series expansion. I think the best Idea possible would be to show the the n'th q-difference equation decays exponentially. I can show the n'th diference equation has exponential decay. I'm not sure how to incorporate that q is the multiplier though. Vladimir Reshetnikov Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 01/14/2017, 09:09 PM (This post was last modified: 01/15/2017, 11:38 PM by Vladimir Reshetnikov.) By the way, in the well-known representation of the tetration as an exponential series ${^z a} = \sum_{m=0}^{\infty} \; c_m q^{mz},$ the coefficients have a q-binomial representation $c_m=\sum_{n=m}^\infty \sum_{k=0}^{n}(-1)^{m+n+k} \; q^{\binom{n-k}{2} + \binom{m}{2}-m(n-1)}\;\frac{\binom{n}{m}_q\;\binom{n}{k}_q}{(q; \; q)_n}\;({^k a})$ The coefficients also satisfy the recurrence $c_m=\frac{\log(a)}{m\left(1-q^{1-m}\right)}\sum_{k=1}^{m-1}kq^{-k}c_{k}c_{m-k},\;\;m>1$ « Next Oldest | Next Newest »

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