06/09/2017, 05:25 PM

(06/09/2017, 01:11 PM)sheldonison Wrote:It's fairly straight forward to go from tetration to pentation in the bounded case. And quite literally, we can do the same thing to get to hexation from pentation. I have a proof by induction that constructs the bounded hyper-operators, it's a little bulky, but it's rather straight forward. I have a nice expression for them, but computing them is rather difficult. I only ever graphed tetration, seeing as graphing pentation would require a higher accuracy of tetration then I was willing to do. And graphing hexation would require a super high accuracy of both pentation and hexation. The formula is recursive and slowly converging. I'm finalizing the paper, just triple checking and making sure all the proofs don't have holes, and it's clear and concise.(06/08/2017, 09:20 PM)JmsNxn Wrote: Really? Does the same happen for pentation? Does it have a real fixed point? Has anybody bothered to go around and actually construct using Kneser's tetration as a base? I mean, if tetration has a repelling real fixed point then it's easy to get to pentation. If pentation has the same thing, it's easy to get hexation...Just a quick overview on pentation for bases>eta. I have not spent any time thinking about hexation for these bases. How far have you gotten in thinking about pentation and hexation for bases<eta?

This is a graph of Tet( 1.63532449671528 ) Tetration bases greater than eta, and less than this base have three real valued fixed points. Tetration bases greater than this base have one real fixed point. The most straightforward way to generate pentation is from the lower fixed point, somewhere between -2 and -1, using that real valued repelling fixed point. Nuinho first pointed out this Tetration base, with a parabolic upper fixed point.

And here is a graph of pentation base(2), from Tet_2's fixed point of -1.743909176132. This pentation base2 graph does not have any real valued fixed points. There are singularities in Pent(z+1) wherever Pent(z) is a negative integer<=-2; this should happen in the complex plane arbitrarily close to the real axis, as z gets bigger.

That's disappointing that pentation has singularities arbitrarily close to the real line. But I've long since had a hunch that getting for would be next to impossible, and one would have to settle with for . Pentation also looks a little wonky, it's got a weird curvature to it. It looks like there could be fixed points, so maybe hexation is possible.

Sadly, how would one show these fixed points are repelling? I have no idea. How would one actually show they exist, I also have no idea. I doubt they exist for all bases though, I wonder if was large enough then pentation doesn't have a fixed point.