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 hyper-operations in a way that isn't recursive, such as to accept non-integers?
Thoughts on hyper-operations of rational but non-integer orders?
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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 hyper-operations in a way that isn't recursive, such as to accept non-integers?
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 math-paper and look for easiness only the lines and their crossings of the coordinate-system 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 coordinate-papers 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 3-D-printer, but I've not yet started to initialize the required data. But I think, that mind-model 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 half-iterate of the superfunction. Regards Tommy1729 (09/01/2019, 04:34 AM)VSO Wrote: Does anyone have any ideas of how to consider hyper-operations in a way that isn't recursive, such as to accept non-integers?Although, each level of the Ackermann function is primitive resursive. The entire Ackermann function itself can not be de-recursed.
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 hyper-operations in a way that isn't recursive, such as to accept non-integers?Although, each level of the Ackermann function is primitive resursive. The entire Ackermann function itself can not be de-recursed. 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 de-recursed. 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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