A relaxed \( \zeta \)-extensions of the Recursive Hyperoperations
I want to show you an easy extension for hyperoperations.
I don't want it to be the most natural, but I want to ask if someone already used this extension and if it can be usefull for something.
Since is a bit different I want to use the plus-notation (\( +_{\sigma} \)) for the hyperoperators.
I start with these basic definitions over the naturals \( b,n,\sigma \in \mathbb{N} \):
\( o)\,\,\,S(n)=n+1 \)
\( o')\,\,\,B_b( \sigma+1):=
\begin{cases}
b, & \text{if} \sigma=0 \\
0, & \text{if} \sigma=1 \\
1, & \text{if} \sigma\gt 1 \\
\end{cases}\)
Then the recursive definitions of the operators \( b,n,\sigma \in \mathbb{N} \)
\( i)\,\,\,b+_0 n=S(n) \)
\( ii)\,\,\,b+_{\sigma+1}0=B_b(\sigma+1) \)
\( iii)\,\,\,b+_{\sigma+1}S(n)=b+_{\sigma}(b+_{\sigma+1}n) \)
Observation before the extension's definitons
\( b+_0 n=1+n \)
\( b+_1 n=b+n \)
we can see that from rank zero to rank one we can define infinite functions \( b+_\sigma n=\zeta_\sigma+n \) with \( 1\lt\zeta_\sigma\lt b \)
\( b+_0 n=\zeta_{0}+n=1+n \)
\( b+_{0.5} n=\zeta_{0.5}+n \)
\( b+_1 n=\zeta_{1}+n=b+n \)
Generalizing, now we can define \( \zeta_b \) as a continous functions from the interval \( 0 \) to \( 1 \), to the interval \( 1 \) to \( b \):
\( Eiv)\,\,\,\zeta_b:[0,1]\rightarrow [1,b] \)
\( Ev)\,\,\,\zeta_b(\varepsilon)=\begin{cases}
1, & \text{if \(\varepsilon=0\)} \\
b, & \text{if \(\varepsilon=1\) } \\ \end{cases} \)
And we can define the operations with fractional rank starting from the interval \( [0,1] \)
\( Evi)\,\,\, b +_{\varepsilon}n=\zeta _b(\varepsilon)+_{1}n \,\, \text{ and} \,\, \varepsilon \in [0,1] \)
Other operations are these ( \( \varepsilon \in ]0,1] \) and \( k \in \mathbb{N} \) ):
\( b +_{k+\varepsilon}n=\zeta _b(\varepsilon)+_{k+1} n \)
Example of \( \zeta _b \) functions and the generated \( \zeta _b \)-hyperoperations:
\( \zeta _b(\varepsilon)=b^\varepsilon \) and \( k \in \mathbb{N} \)
for \( \varepsilon \in ]0,1] \) and \( k \in \mathbb{N} \)
\( b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n \)
I want to show you an easy extension for hyperoperations.
I don't want it to be the most natural, but I want to ask if someone already used this extension and if it can be usefull for something.
Since is a bit different I want to use the plus-notation (\( +_{\sigma} \)) for the hyperoperators.
I start with these basic definitions over the naturals \( b,n,\sigma \in \mathbb{N} \):
\( o)\,\,\,S(n)=n+1 \)
\( o')\,\,\,B_b( \sigma+1):=
\begin{cases}
b, & \text{if} \sigma=0 \\
0, & \text{if} \sigma=1 \\
1, & \text{if} \sigma\gt 1 \\
\end{cases}\)
Then the recursive definitions of the operators \( b,n,\sigma \in \mathbb{N} \)
\( i)\,\,\,b+_0 n=S(n) \)
\( ii)\,\,\,b+_{\sigma+1}0=B_b(\sigma+1) \)
\( iii)\,\,\,b+_{\sigma+1}S(n)=b+_{\sigma}(b+_{\sigma+1}n) \)
Observation before the extension's definitons
\( b+_0 n=1+n \)
\( b+_1 n=b+n \)
we can see that from rank zero to rank one we can define infinite functions \( b+_\sigma n=\zeta_\sigma+n \) with \( 1\lt\zeta_\sigma\lt b \)
\( b+_0 n=\zeta_{0}+n=1+n \)
\( b+_{0.5} n=\zeta_{0.5}+n \)
\( b+_1 n=\zeta_{1}+n=b+n \)
Generalizing, now we can define \( \zeta_b \) as a continous functions from the interval \( 0 \) to \( 1 \), to the interval \( 1 \) to \( b \):
\( Eiv)\,\,\,\zeta_b:[0,1]\rightarrow [1,b] \)
\( Ev)\,\,\,\zeta_b(\varepsilon)=\begin{cases}
1, & \text{if \(\varepsilon=0\)} \\
b, & \text{if \(\varepsilon=1\) } \\ \end{cases} \)
And we can define the operations with fractional rank starting from the interval \( [0,1] \)
\( Evi)\,\,\, b +_{\varepsilon}n=\zeta _b(\varepsilon)+_{1}n \,\, \text{ and} \,\, \varepsilon \in [0,1] \)
Other operations are these ( \( \varepsilon \in ]0,1] \) and \( k \in \mathbb{N} \) ):
\( b +_{k+\varepsilon}n=\zeta _b(\varepsilon)+_{k+1} n \)
Example of \( \zeta _b \) functions and the generated \( \zeta _b \)-hyperoperations:
\( \zeta _b(\varepsilon)=b^\varepsilon \) and \( k \in \mathbb{N} \)
for \( \varepsilon \in ]0,1] \) and \( k \in \mathbb{N} \)
\( b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n \)
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)