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 \( k \in \mathbb{N} \)

\( b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n \)

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)