A new way of approaching fractional hyper operators - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: A new way of approaching fractional hyper operators (/showthread.php?tid=737) A new way of approaching fractional hyper operators - JmsNxn - 05/26/2012 I've figured out a new approach for solving hyper operators; for finding a smooth curve from addition to exponentiation for all natural numbers. It's a rather tricky method and I'm still a little shaky on the details. Firstly, I considered that the operators cannot be commutative/associative over an interval because that implies discontinuity at at least exponentiation. However, they can be commutative/associative at single points. Anyway, the juxt of the idea is by exploiting the fundamental theorem of arithmetic. We start with: $x = \prod_i \rho_i^{\mu_{x;i}}$ And create a function $\psi$ such that: $\psi(x) = \sum_i \mu_{x;i} \rho_i$ which has the following properties: $\psi(\rho) = \rho$ $\psi(xy) = \psi(x) + \psi(y)$ But also, since $\psi(x)$ is a natural number; it has a prime decomposition: $\psi(x) = \prod_i \rho_i^{\kappa_{x;i}}$ Essentially then, we will take points inbetween addition and multiplication by transforming $\mu_{x;i}$ into $\kappa_{x;i}$. Or, writing more mathematically. If $x \otimes y$ is an operator between multiplication and addition, and $x^{\otimes y}$ is its super operator, and this is one of the operators which happen to be commutative/associative; then we create a one to one function $\lambda$ with the following properties: $\lambda(\lambda^{-1}(x)\lambda^{-1}(y)) = x \otimes y$ $\lambda(\lambda^{-1}(x)^y) = x^{\otimes y}$ where $\lambda(x) = \prod_i \rho_i^{\eta_{x;i}}$ where here $\eta_{x;i}$ is a real inbetween $\kappa_{x;i}$ and $\mu_{x;i}$. and therefore: $\lambda(\rho) = \rho$ adding the requirement the identity function $\pi$ is $\pi(s + 1) = 1$ if $x[s]y = x \otimes y$. These operators satisfy the recursion formula: $x \otimes x^{\otimes y} = x^{\otimes y + 1}$ This gives us a very beautiful commutative operator which has a lot of beautiful properties that I haven't written out. However, we cannot have an interval of these operators or else the end product will not be analytic. Therefore the question is, how to create an interval of non-associative, non-commutative operators which fill the gaps between these commutative operators. Also, what operators in between addition and multiplication should qualify as a commutative operator? I've taken to the knack of calling these commutative and associative operators as stars. Once we have a full uniqueness criterion for them: We would then, figuratively speaking, draw a line between the discrete transformations of stars filling the gaps with non-commutative/non-associative operators. This technique should work for operators greater than addition and less than and including exponentiation. Specifically we can evaluate any commutative operation above addition and below multiplication at any two primes, or a number with a single prime factor; and we can evaluate the same numbers above multiplication and below exponentiation for natural values: $\rho^{\otimes m} \otimes \rho^{\otimes n}$ where $\otimes = [s]$ and $^{\otimes} = [s + 1]$. We stress that we only evaluate $s$ as a complex variable. I hope to develop some techniques for creating the function $\lambda$ in a way such that we can evaluate for specific operators of order $s$; and we have a uniqueness as to why we use this function for a specific value of $s$. Any help or comments would be greatly appreciated. Or maybe I'm doing something awfully wrong! Thanks.