• 2 Vote(s) - 3 Average
• 1
• 2
• 3
• 4
• 5
 Andrew Robbins' Tetration Extension bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/07/2007, 04:38 PM (This post was last modified: 08/23/2009, 02:55 PM by bo198214.) I just read Andrew Robbins' solution to the tetration problem, which I find very convincing, and want to use the opportunity to present and discuss it here. The solution ${}^y x$ satisfies the 2 natural conditions 1. ${}^1b=b$ and ${}^{x+1}b=b^{{}^xb}$ 2. $x\mapsto b^x$ is infinitely differentiable. However for avoiding difficulties with a later expansion, he instead solves for the tetration logarithm tlog (which he calls super logarithm but I find "tetration logarithm" somewhat more specific), which is the inverse of $x\mapsto b^x$, i.e. ${}^{\text{tlog}_b(x)}b=x$. The first condition is then translated into 1'. $\text{tlog}_b b=1$ and $\text{tlog}_b(x)=\text{tlog}_b(b^x)-1$ while the second condition is equivalent to that also 2'. $x\mapsto \text{tlog}_b(x)$ is infinitely differentiable. By 1' we merely need to consider the tlog on the interval $(0,1\]$ and can then derive the values for the other intervals $(1,b\]$, $(b,b^b\]$, $(b^b,b^{b^b}\]$ etc. above and $(-\infty,0\]$ below. The idea is now that we define a smooth (infinitely differentiable) function t on $(0,1\]$ and ensure that the by 1' resulting function is also smooth at the joining points of the intervals. Obviously it suffices to ensure this for the joining point 0 (and by 1' this transposes to each other joining point). We simply expand t into a power series at 0 and then try to determine the coefficients such that the resulting function is smooth at 0 $t(x) = \sum_{i=0}^\infty \nu_i \frac{x^i}{i!}$ $\nu_i$ is the i-th derivative of t at 0. The resulting function on $(-\infty,0\]$ is $s(x)=t(b^x)-1$ We have to ensure that $s^{(k)}(0)=t^{(k)}(0)=\nu_k$ for each $k\ge 0$. What is now the k-th derivative of s at 0? For $k=0$ we get $\nu_0=s(0)=t(b^0)-1=t(1)-1=\text{tlog}_b(b^1)-2=\text{tlog}_b(b)-2=1-2=-1$. For $k\ge 1$ the constant -1 vanishes and we make the following calculations: $s^{(k)}(x)=(t(b^x))^{(k)}=\left(\sum_{i=0}^\infty \nu_i \frac{b^{xi}}{i!}\right)^{(k)}=\sum_{i=0}^\infty\frac{\nu_i}{i!}(b^{xi})^{(k)}$ The derivation of $b^{xi}$ is easily determined to be $(b^{xi})'=b^{xi}\text{ln}(b) i$ and so the k-th derivative is $(b^{xi})^{(k)} = b^{xi}(\text{ln}(b)i)^k$, which give us in turn $\nu_k=s^{(k)}(0)= \text{ln}(b)^k\sum_{i=0}^\infty\nu_i\frac{i^k}{i!}$ for $k\ge 1$. This is an infinite linear equation system system in the variables $\nu_1,\nu_2,\dots$. The way of Andrew Robbins is now to approximate a solution by considering finite linear equation systems consisting of n equations and n variables $\nu_{i,n}$ resulting from letting $\nu_{k,n}=0$ for $k>n$. First one can show that these equation systems have a unique solution for b>1 and numerical evidence then shows that $\nu_k=\lim_{n\to\infty}\nu_{k,n}$ converges and that the resulting $\nu_k$ are a solution of the infinite equation system. Further numerical evidence shows, that the infinite sum in the definition of t converges for the so obtained $\nu_i$. However I would guess that the claimed uniqueness for a solution satisfying 1' and 2' is not guarantied. We can use different approximations, for example for a given constant we can consider the equation systems, resulting from letting $\nu_k=c$ for $k>n$. Because interestingly the sum $\sum_{i=0}^\infty \frac{i^k}{i!}$ converges to $eB_k$ where e is the Euler constant and B_k are the Bell numbers. So by setting $\nu_k=c$ for $k>n$ the remaining sum $\sum_{i=n+1}^\infty c\frac{i^k}{i!}$ converges and merely introduces an additive constant in the linear equation system. The obtained solutions are different from the solution obtained by c=0. However I didnt verify yet the convergence properties of these alternative solutions. « Next Oldest | Next Newest »

 Messages In This Thread Andrew Robbins' Tetration Extension - by bo198214 - 08/07/2007, 04:38 PM RE: Andrew Robbins' Tetration Extension - by bo198214 - 08/18/2007, 08:20 PM RE: Andrew Robbins' Tetration Extension - by bo198214 - 08/19/2007, 09:50 AM RE: Andrew Robbins' Tetration Extension - by bo198214 - 08/20/2007, 02:22 PM RE: Andrew Robbins' Tetration Extension - by andydude - 11/12/2007, 08:43 AM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 06/26/2009, 10:51 PM RE: Andrew Robbins' Tetration Extension - by bo198214 - 06/27/2009, 09:39 AM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 06/28/2009, 12:08 AM RE: Andrew Robbins' Tetration Extension - by jaydfox - 11/06/2007, 04:17 AM RE: Andrew Robbins' Tetration Extension - by jaydfox - 11/06/2007, 04:27 AM RE: Andrew Robbins' Tetration Extension - by bo198214 - 11/06/2007, 10:57 AM RE: Andrew Robbins' Tetration Extension - by jaydfox - 11/06/2007, 01:58 PM RE: Andrew Robbins' Tetration Extension - by bo198214 - 11/06/2007, 03:58 PM RE: Andrew Robbins' Tetration Extension - by jaydfox - 11/12/2007, 09:14 AM RE: Andrew Robbins' Tetration Extension - by andydude - 11/12/2007, 09:56 AM RE: Andrew Robbins' Tetration Extension - by bo198214 - 11/12/2007, 08:05 PM RE: Andrew Robbins' Tetration Extension - by andydude - 11/13/2007, 12:16 AM RE: Andrew Robbins' Tetration Extension - by bo198214 - 11/13/2007, 10:21 AM RE: Andrew Robbins' Tetration Extension - by andydude - 11/13/2007, 05:45 PM RE: Andrew Robbins' Tetration Extension - by Gottfried - 03/17/2008, 07:52 AM RE: Andrew Robbins' Tetration Extension - by Gottfried - 03/17/2008, 06:09 PM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 06/29/2009, 08:20 PM RE: Andrew Robbins' Tetration Extension - by andydude - 07/27/2009, 08:10 AM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 08/11/2009, 12:18 PM RE: Andrew Robbins' Tetration Extension - by jaydfox - 08/11/2009, 07:06 PM RE: Andrew Robbins' Tetration Extension - by jaydfox - 08/11/2009, 07:12 PM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 08/23/2009, 02:45 PM RE: Andrew Robbins' Tetration Extension - by bo198214 - 08/23/2009, 03:23 PM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 08/26/2009, 04:01 PM RE: Andrew Robbins' Tetration Extension - by andydude - 09/04/2009, 06:42 AM RE: Andrew Robbins' Tetration Extension - by Gottfried - 12/28/2009, 05:21 PM RE: Andrew Robbins' Tetration Extension - by tommy1729 - 08/18/2016, 12:29 PM RE: Andrew Robbins' Tetration Extension - by Gottfried - 08/22/2016, 04:19 PM

 Possibly Related Threads... Thread Author Replies Views Last Post Possible continuous extension of tetration to the reals Dasedes 0 1,678 10/10/2016, 04:57 AM Last Post: Dasedes Non-trivial extension of max(n,1)-1 to the reals and its iteration. MphLee 3 4,631 05/17/2014, 07:10 PM Last Post: MphLee extension of the Ackermann function to operators less than addition JmsNxn 2 4,760 11/06/2011, 08:06 PM Last Post: JmsNxn Tetration Extension to Real Heights chobe 3 7,063 05/15/2010, 01:39 AM Last Post: bo198214 Tetration extension for bases between 1 and eta dantheman163 16 22,365 12/19/2009, 10:55 AM Last Post: bo198214 Extension of tetration to other branches mike3 15 24,438 10/28/2009, 07:42 AM Last Post: bo198214 andrew slog tommy1729 1 3,921 06/17/2009, 06:37 PM Last Post: bo198214 Dmitrii Kouznetsov's Tetration Extension andydude 38 44,169 11/20/2008, 01:31 AM Last Post: Kouznetsov Hooshmand's extension of tetration andydude 9 11,984 08/14/2008, 04:48 AM Last Post: Danesh

Users browsing this thread: 1 Guest(s)