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 2^^^1.5 = ? robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 04/05/2020, 05:29 PM Since 2Hn1 and 2Hn2 are fixed points at 2 and 4 I thought I'd experiment a little with 2Hn1.5 and curiously it seems to approach e, going from 3.5 to 3 to 2.828 to 2.745 as n increases. I'm curious to see if the trend continues with n=5, 2^^^1.5. I know it can be expressed as x when x^^x = 65536 but beyond that I'm stumped, anyone have any idea how to solve such an equation? Daniel Fellow Posts: 70 Threads: 25 Joined: Aug 2007 04/10/2020, 06:58 AM Sorry if I don't want to fight this war one number at a time. I believe that the most effective approach to the higher hyperoperators for problems like $2\uparrow{^k} 1.5$ is to tackle the issue for all smooth functions $f^t(x)$. This then gives a solution for tetration, pentation and all higher hyperoperators.  http://tetration.org/Tetration/index.html https://www.overleaf.com/read/zjwkzgftsqkm sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 04/10/2020, 08:56 PM (04/05/2020, 05:29 PM)robo37 Wrote: Since 2Hn1 and 2Hn2 are fixed points at 2 and 4 I thought I'd experiment a little with 2Hn1.5 and curiously it seems to approach e, going from 3.5 to 3 to 2.828 to 2.745 as n increases. I'm curious to see if the trend continues with n=5, 2^^^1.5. I know it can be expressed as x when x^^x = 65536 but beyond that I'm stumped, anyone have any idea how to solve such an equation? $2\uparrow\uparrow 1.5$ is tetration or iterated eexponentaton; Kneser's solution gives 2^^1.5~=2.7448 $2\uparrow\uparrow\uparrow 1.5$ requires analytic pentation, or iterated tetration, or  isn't as well behaved or as uniquely defined, but using the lower fixed point of tetration base 2; Tet_2(-1.7439)=~-1.7439 gives an analalytic pentation for base 2, and then 2^^^1.5~=2.7029 There is also an analytic hexation function; 2^^^^1.5=~2.6729 fatou.gp implements it. - Sheldon - Sheldon « Next Oldest | Next Newest »

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