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 [2015] 4th Zeration from base change pentation tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 03/24/2015, 01:22 PM Recently I posted a 3rd zeration and now Im considering a 4th. It may be that the 3rd equals one of the first 2 and that the 4th is also equal to another. Anyway here is the logic. the base or unit satisfies approximately : b[3]x = b^x => base change => x -> C * x = C[2]x b[4]x = b^^x => base change => x -> C + x = C[1]x b[5]x => base change => x -> C[0]x or C[0]x. So from a pentation base change we get zeration. Well maybe. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/24/2015, 10:45 PM Sorry bu how works tetration's base change? MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 03/28/2015, 09:04 PM Our forum has a search option. Anyway I think this thread is very good in explaining : http://math.eretrandre.org/tetrationforu...=14&page=5 At that time I was a newbie here so forgive any silly ideas regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/29/2015, 03:51 PM Ah thanks... I'm a newbie now so I can't judge you old ideas. Actually I don't even understand how Jay came at that formula and reading bo seems to me that the equation is not proved yet. Anyways, assuming it is correct, it looks amazing! It wold be really amazing to find that the basechange of pentation is linked with something that is in some sense "before the addition". But the question is ... in wich sense is "before the addition"? In fact, as long as we want the "Akermann" recurrence (call it mother law or how you like) to hold across the ranks (negative and non-natural as well) we are still with the problem of the successor function being a fixpoint of the subfunction operator... aka all the negative ranks are the successor (as Bo noted long ago, as Jmsnxn proved and as you noted long ago in JmsNxn's post about extending Ackermann). Also if we drop the "Ackermann's recursion" ($H_{s+1}\circ S=H_{s}\circ H_{s+1}$) then we just have to move on other Hyperoperations families... like Bennet's one... but imho Ackermann's recursion is too much important and fascinating, if you think about it is that recursion that give us tetration and pentation, ..., and $s{\rm -ation}$, while in other families like the balanced, the bennet or the right-distributive (lower Hos) we never reach them. MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 03/29/2015, 05:03 PM (03/29/2015, 03:51 PM)MphLee Wrote: Ah thanks... I'm a newbie now so I can't judge you old ideas. Actually I don't even understand how Jay came at that formula and reading bo seems to me that the equation is not proved yet. Anyways, assuming it is correct, it looks amazing! It wold be really amazing to find that the basechange of pentation is linked with something that is in some sense "before the addition". But the question is ... in wich sense is "before the addition"? In fact, as long as we want the "Akermann" recurrence (call it mother law or how you like) to hold across the ranks (negative and non-natural as well) we are still with the problem of the successor function being a fixpoint of the subfunction operator... aka all the negative ranks are the successor (as Bo noted long ago, as Jmsnxn proved and as you noted long ago in JmsNxn's post about extending Ackermann). Also if we drop the "Ackermann's recursion" ($H_{s+1}\circ S=H_{s}\circ H_{s+1}$) then we just have to move on other Hyperoperations families... like Bennet's one... but imho Ackermann's recursion is too much important and fascinating, if you think about it is that recursion that give us tetration and pentation, ..., and $s{\rm -ation}$, while in other families like the balanced, the bennet or the right-distributive (lower Hos) we never reach them. The equation has been proven to be correct. ALthough I was not able to find that particular post here , im pretty sure it exists here. The main idea is that things like ln(ln(ln( ... 3^3^3^ ..))) converge rapidly. Im aware of the ackermann recursion and your related comments. But there is as you noted , no other choice. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 03/29/2015, 05:47 PM In case it was not clear , I think the zeration will be the one that satisfies 2[0]2[0] ... = x + 2 or this one a[0]b = ln(exp(a)+exp(b) ). The ackermann recursion makes no sense in this context for pentation. Numerical experiments seem hard at first because of the speed of pentation. And I have 0 theory. But maybe others have ideas. regards tommy1729 « Next Oldest | Next Newest »

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