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 Zeration reconsidered using plusation. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/29/2015, 08:07 AM X^^1 = x X^1 = x X*1 = x X(+)1 = x (+) is plusation. It came to me that zeration ideas failed or were nonintresting because of using simple addition. Therefore i now consider this plusation which is more logical. X(+)2 = x + 1 For c real X(+)(2^c) = x + c. = x +1 (c times) Notice x * 2^y = x + x (y times) X ^ 2^y = x^2 (y times) So plusation is a Natural logical operator in the list of noncommutative hyperoperators. Plusation is like the superfunction of zeration. Regards Tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/23/2015, 03:39 PM (This post was last modified: 10/23/2015, 04:56 PM by MphLee.) It is not exactly the supefunction of zeration (max-kroenecker delta definition)...but it is the superfunction of Bennett's base 2 preaddition $\odot_{-1}$ (-1th rank in base 2 commutative hos hierarchy). $A\odot_{i} B=\exp_2^{\circ i}[\log_2^{\circ i}(A) +\log_2^{\circ i}(B)]$ $A\odot_0 B=A+B$ $A\odot_{-1} B=\log_2(2^A +2^B)$ In fact you are right: define plusation (set it as rank 1 of a new sequence) $b(+)_1x=b+\log_2(x)$ $b(-)_1 x=2^{x-b}$ Let's take it's subfunction in the variable x (i.e.$F\mapsto f$ where $F(x+1)=f(F(x))$ ) $b(+)_0x=b(+)_1(1+b(-)_1 x)=\log_2(2^{x}+2^b)=b\odot_{-1}x$ But the sequence $(+)_t$ doesn't seem much interesting or natural imho... the only nice properties are that 1) it is based on the usual recursion/iteration law (ML $b*_{i+1}x+1=b*_i(b*_{i+1}x)$ ) 2) it intersects the 2-based commutative hos sequence at t=0 ($b(+)_0x=b\odot_{-1}x$) MathStackExchange account:MphLee « Next Oldest | Next Newest »

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