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Say we define using hyperoperators with non-integers the same way we do with exponentiation, that is to say we convert the 'exponent' to a fraction and raise the base to the hyper-power of the numerator, then taking it to the hyper-root of the denominator.

What would this make 2^^^1.5? That is 2^^^(3/2) = 2nd pent-root of 2^^^3 = value of x when x^^x = 65536?

I ask because obviously using 1 and 2 as hyper-powers of 2 are fixed points at 2 and 4 so 1.5 must be bounded at some value in between, and solving for the first 4 hyper-operators brings us 3.5, 3, 2.828 and 2.745 ~ suspiciously slowing down as it approaches e don't you think? So I'm curious if solving for pentation will continue the trend and close in on e.

Alternatively, estimating e^^e and seeing if it falls anywhere near 65536 could prove interesting and we might be able to use it to prove if 2^^^1.5 is therefore greater than or less than e.

Dakirel
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  • Here's a MathJax tutorial :) – Shaun Apr 08 '20 at 19:55
  • Short answer: No, your definition isn't even self-consistent e.g. $a\uparrow\uparrow\uparrow4\ne(a\uparrow\uparrow\uparrow2)\uparrow\uparrow\uparrow2$ for general $a$. The fact that $a^{m/n}=\sqrt[n]{a^m}$ relies on the fact that multiplication is commutative, which exponentiation and tetration lack, so there's no reason to expect this to hold for tetration or pentation. – Simply Beautiful Art Apr 09 '20 at 00:50
  • Okay but it seemed to work with tetration, throwing x^x=16 into WolframAlpha yields the result x ≈ 2.7453680235674634847 regardless of whether or not 2^^(3/2) is well defined or commutative. Why can't we do the same for pentation? – Dakirel Apr 09 '20 at 17:59
  • It doesn't work for tetration either because $x\widehat~x\widehat~x\widehat~x\ne(x\widehat~x)\widehat~(x\widehat~x)$ i.e. $x\widehat~\widehat~4\ne(x\widehat~\widehat~2)\widehat~\widehat~2$. – Simply Beautiful Art Apr 12 '20 at 15:58

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