• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Recursive formula generating bounded hyper-operators JmsNxn Long Time Fellow Posts: 465 Threads: 85 Joined: Dec 2010 01/17/2017, 05:10 AM (This post was last modified: 01/17/2017, 05:14 AM by JmsNxn.) Hello everybody. So I've come across the following formula for all hyper-operators with base $1 < \alpha < e^{1/e}$. There are a few holes in my proof of how to construct these functions, but the holes are more than anything superficial, and in fact have more to do with showing inconsequential details of the solution. The formulas all work out. Consider the following construction. Let us start with $\alpha \uparrow^0 z = \alpha \cdot z$ and define $\alpha \uparrow^n z$ recursively using the following formulas $\alpha \uparrow^{n} z = \sum_{k=0}^\infty \binom{z}{k}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\alpha \uparrow^{n-1}\alpha\uparrow^{n-1}...(j\,times)...\uparrow^{n-1}\alpha$ So that exponentiation equals $\alpha^z =\sum_{k=0}^\infty\binom{z}{k}\sum_{j=0}^k (-1)^{k-j} \binom{k}{j}\alpha^j =\sum_{k=0}^\infty \binom{z}{k}(\alpha - 1)^k$ which follows from the usual binomial formula. And then tetration (which is well known, this I can prove absolutely without a doubt) $\alpha \uparrow^2 z = (^z \alpha) =\sum_{k=0}^\infty \binom{z}{k}\sum_{j=0}^k (-1)^{k-j} \binom{k}{j}(^j \alpha)$ But not so obviously, and definitely not through little work, this works for pentation $\alpha \uparrow^{3} z = \sum_{k=0}^\infty \binom{z}{k}\sum_{j=0}^k (-1)^{k-j} \binom{k}{j}\alpha \uparrow^2 \alpha \uparrow^2 ...(j\,times)...\uparrow^2 \alpha$ This construction works using a proof by induction. There are some cracks in the proof so far, but nonetheless everything is in working order. The only missing pieces are to do with some problems in integral calculus. I can state them here for working order, if someone is curious. I could elaborate on how these functions are constructed, but if one is interested, the entire construction, in its full working order is here https://arxiv.org/pdf/1503.07555.pdf Minus the simple proof that these functions can be expanded in a newton series. Essentially fivexthethird gave the basis of this proof, since then I've made it more rigorous, he talks in the case of one example, but it works in the case of all bounded functions http://math.eretrandre.org/tetrationforu...ht=tpid+13    I must note, the problem with the paper I posted is in fact rather minute. The paper is a little disorganized, I'm trying to complete it right now. The major problem is arguing more rigorously why some of the integral transforms do in fact converge, of which I'm in the process of finalizing. Nonetheless the most important thing I've been asking myself lately is something really rather simple, it has to do with the Newton construction. The exact functions I have defined. Does $\alpha \uparrow^n z$ have an imaginary period $\ell$ such that $\alpha \uparrow^n z + i \ell = \alpha \uparrow^n z$ for all $1 < \alpha < \eta$? Do the hyper-operators with these bases always have an imaginary period? For tetration and for exponentiation this follows trivially. It is something inherent to their definition. What is great about this question though, is that it has many incarnations, it extends very far. If $\alpha \uparrow^n \infty = \omega_{n}$ then this question can be rephrased as: is $\sum_{k=0}^\infty |\alpha \uparrow^{n} k - \omega_n| < \infty$? This question can also be rephrased as if $F(z) = \frac{d}{dz}\alpha\uparrow^{n-1} z$, is $F(\omega_n) < 1$ for all $1 <\alpha < \eta$? and lastly: If $\alpha \uparrow^{n} \omega_\alpha = \omega_\alpha$ and $\frac{d}{d\alpha}\alpha\uparrow^n x > 0$ then is $\omega_\alpha$ analytic for all $1 < \alpha < \eta$? I could restate this question in many ways, but it gives me no advantage at showing it. Therefore I am stuck. And I call upon somebody to at least give me numerical evidence that these functions generated through Newton series are in fact periodic with imaginary period. That pentation in the interval $1 < \alpha < \eta$ is periodic with imaginary period. That hexation in the interval $1 < \alpha < \eta$ is periodic with imaginary period. I believe this is a great question about the regularity of iteration. That in fact every hyper operator with base $1 < \alpha < \eta$ has an imaginary period (at least when using the standard Schroder iteration). « Next Oldest | Next Newest »

 Messages In This Thread Recursive formula generating bounded hyper-operators - by JmsNxn - 01/17/2017, 05:10 AM

 Possibly Related Threads... Thread Author Replies Views Last Post On my old fractional calculus approach to hyper-operations JmsNxn 13 562 05/29/2021, 01:13 AM Last Post: MphLee hyper 0 dantheman163 2 4,650 03/09/2021, 10:28 PM Last Post: MphLee On to C^\infty--and attempts at C^\infty hyper-operations JmsNxn 11 1,992 03/02/2021, 09:55 PM Last Post: JmsNxn There is a non recursive formula for T(x,k)? marraco 5 1,901 12/26/2020, 11:05 AM Last Post: Gottfried Thoughts on hyper-operations of rational but non-integer orders? VSO 2 3,050 09/09/2019, 10:38 PM Last Post: tommy1729 Hyper-volume by integration Xorter 0 2,642 04/08/2017, 01:52 PM Last Post: Xorter Hyper operators in computability theory JmsNxn 5 8,334 02/15/2017, 10:07 PM Last Post: MphLee Is bounded tetration is analytic in the base argument? JmsNxn 0 2,669 01/02/2017, 06:38 AM Last Post: JmsNxn Extrapolated Faá Di Bruno's Formula Xorter 1 3,906 11/19/2016, 02:37 PM Last Post: Xorter Rational operators (a {t} b); a,b > e solved JmsNxn 30 61,707 09/02/2016, 02:11 AM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)