09/01/2019, 04:34 AM
I can't seem to find the right angle to approach this concept intuitively. Does anyone have any ideas of how to consider hyperoperations in a way that isn't recursive, such as to accept nonintegers?
Thoughts on hyperoperations of rational but noninteger orders?

09/01/2019, 04:34 AM
I can't seem to find the right angle to approach this concept intuitively. Does anyone have any ideas of how to consider hyperoperations in a way that isn't recursive, such as to accept nonintegers?
I am not sure I get your problem correctly.
Take the function \( f: b^z \) as to be iterated, with, say \( b=sqrt(2) \) . Assume one plane on a mathpaper and look for easiness only the lines and their crossings of the coordinatesystem of integer complex numbers \( z_0 \) . Now take another paper, position it 10 cm above and for every point of the crossings (and ideally also of the lines) mark the values of \( z_1=b^{z_0} \). Then repeat it with a third plane, again 10 cm above, marking \( z_2=b ^{b^{z_0}} \) . After that, try to connect the related points of the zero'th, the first and the second plane by a weak string, say a spaghetti or so. Surely except of the fixpoints in \( z_0 \) it shall be difficult to make a meaningful and smooth curve  and in principle it seems arbitrary, except at the fixpoints, where we simple stitch a straight stick through the iterates of the \( z_0 \) at the fixpoint. Of course the spaghetti on the second level is then no more arbitrary but must be  point for point  be computed by one iteration. But the spaghatti in the first level follow that vertically orientated curve, where a fictive/imaginative plane of paper is at fractional heights and the fractional iterates would be the marks on the coordinatepapers at the "fractional (iteration) height". I'd liked to construct some physical example, showing alternative paths upwards between the fixed basic planes, with matrial curves made by an 3Dprinter, but I've not yet started to initialize the required data. But I think, that mindmodel alone makes it possibly already sufficiently intuitive for you. A somewhat better illustration is in my answer at MSE, see https://math.stackexchange.com/a/451755/1714
Gottfried Helms, Kassel
09/09/2019, 10:38 PM
I think the OP refers to concepts like , what i called " semi super " operators.
Like the semisuper operator of the semisuper operator of f(x) is the super of f(x). This is extremely difficult. Do not confuse with the functional halfiterate of the superfunction. Regards Tommy1729 (09/01/2019, 04:34 AM)VSO Wrote: Does anyone have any ideas of how to consider hyperoperations in a way that isn't recursive, such as to accept nonintegers?Although, each level of the Ackermann function is primitive resursive. The entire Ackermann function itself can not be derecursed.
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
06/30/2022, 11:41 PM
(06/28/2022, 08:33 AM)Catullus Wrote:(09/01/2019, 04:34 AM)VSO Wrote: Does anyone have any ideas of how to consider hyperoperations in a way that isn't recursive, such as to accept nonintegers?Although, each level of the Ackermann function is primitive resursive. The entire Ackermann function itself can not be derecursed. This comment doesn't address VSO's question. Ackermann function do not coincides with hyperoperations. Secondly, hyperoperations need not to be defined recursively. Third point: ackermann function, goodstein function (natural hyperoperations), if considered as functions over the natural numbers are recursive. Are defined recursively. It is not fully clear what do you mean by derecursed. Anyways, if the question is if it's possible to give an analytical, non recursive, expression to the ackermann function/hyperoperations, the answer is yes. JmsNxn gave one expression for that. The real question is if that analytical representation has desired properties: does it satifies a functional equation? Is it smooth? Holomorphic? MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\) 
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