\( s \)-rank hyperoperations have meaning as long as we can iterate \( s \) times a function \( \Sigma \) defined in the set of the binary functions over the naturals numbers (or defined over a set of binary functions.)
let me explain why.
There are many differente Hyperoperations sequences, end they are all defined in a different way:
we start with an operation \( * \) and we obtain its successor operation \( *' \) applying a procedure \( \Sigma \) (usually a recursive one).
\( \Sigma(*)=*' \)
So every Hyperoperation sequence is obtained applying that recursive procedure \( \Sigma \) to a base operation \( * \) (aka the first step of the sequence)
\( *_0:=* \)
\( *_1:=\Sigma(*_0) \)
\( *_2:=\Sigma(\Sigma(*_0)) \) and so on
or in a formal way
\( *_0:=* \)
\( *_{n+1}:=\Sigma(*_n) \)
That is the same as
\( *_{n}:=\Sigma^{\circ n}(*_{0}) \)
so if we can extend the iteration of \( \Sigma^{\circ n} \) from \( n \in \mathbb{N} \) to the real-complex numbers the work is done.
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let me explain why.
There are many differente Hyperoperations sequences, end they are all defined in a different way:
we start with an operation \( * \) and we obtain its successor operation \( *' \) applying a procedure \( \Sigma \) (usually a recursive one).
\( \Sigma(*)=*' \)
So every Hyperoperation sequence is obtained applying that recursive procedure \( \Sigma \) to a base operation \( * \) (aka the first step of the sequence)
\( *_0:=* \)
\( *_1:=\Sigma(*_0) \)
\( *_2:=\Sigma(\Sigma(*_0)) \) and so on
or in a formal way
\( *_0:=* \)
\( *_{n+1}:=\Sigma(*_n) \)
That is the same as
\( *_{n}:=\Sigma^{\circ n}(*_{0}) \)
so if we can extend the iteration of \( \Sigma^{\circ n} \) from \( n \in \mathbb{N} \) to the real-complex numbers the work is done.
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Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
