Negative, Fractional, and Complex Hyperoperations KingDevyn Junior Fellow Posts: 2 Threads: 2 Joined: May 2014 05/30/2014, 06:58 AM Is there a way to continue the patterns we see within the natural numbers of current hyper-operations (Hyper-1, Hyper-2, Hyper-3, Hyper-4, ect...) or at least prove that we cannot extend the value of the operation to fractional numbers? E.g. Hyper-1/2. Negative numbers? E.g. Hyper-(-2) Or even imaginary numbers? E.g. Hyper-3i. They need not be defined, but are these operations technically there, just without practical use? Or are our names for the hyper-operations strictly for listing and naming purposes, with no way to derive meaning from such a number? Could a fractional, or negative hyper-operation represent an operator we have already defined? E.g. Hyper-(-2)= Division, or Hyper-1/2 = Division? Comments on the controversy of Zeration are also encouraged. Thanks! MphLee Long Time Fellow Posts: 368 Threads: 28 Joined: May 2013 05/30/2014, 07:57 AM (This post was last modified: 05/30/2014, 07:58 AM by MphLee.) $s$-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. ---------------------- MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 368 Threads: 28 Joined: May 2013 05/30/2014, 08:19 AM (This post was last modified: 05/30/2014, 08:22 AM by MphLee.) I'm not sure but I think that bo198214(Henrik Trappmann) had this idea in 2008 http://math.eretrandre.org/tetrationforu...l+function With his idea we can reduce the problem of real-rank hyperoperations to an iteration problem Later this idea was better developed by JmsNxn (2011) with the concept of "meta-superfunctions" http://math.eretrandre.org/tetrationforu...hp?tid=708 I'm still working on his point of view but there is a lot of work to do... MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »

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