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2^^^1.5 = ?
#1
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?
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#2
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  is to tackle the issue for all smooth functions . This then gives a solution for tetration, pentation and all higher hyperoperators. 
http://tetration.org/Tetration/index.html

https://www.overleaf.com/read/zjwkzgftsqkm
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#3
(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?

is tetration or iterated eexponentaton; Kneser's solution gives 2^^1.5~=2.7448
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
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