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 A relaxed $$\zeta$$-extensions of the Recursive Hyperoperations MphLee Fellow Posts: 177 Threads: 16 Joined: May 2013 06/14/2013, 09:57 PM (This post was last modified: 06/15/2013, 08:28 AM by MphLee.) 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 $k \in \mathbb{N}$ $b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n$ MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ « Next Oldest | Next Newest »

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