exp(x) - 1 tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/14/2009, 11:48 PM this idea might have been posted before, forgive me if such is the case. we know exp(x)-1 has a fixed point. which leads to unique half-iterate and a somewhat unique superfunction. so we might want to use the fixpoint of exp(x) - 1 for a ' surrogate fixpoint ' of exp(x). here is how - if i dont blunder - : using superfunction F(x) : F(x + 1) = exp [ F(x) ] - 1. which can be solved by taylor series i believe ? now the simple but brilliant idea - if correct - F( x + 1 ) + 1 = exp [ F(x) ] generalize to F ( x + a ) + a = exp exp exp ... a times [ F(x) ] and Coo tetration follows !!? regards tommy1729 Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/15/2009, 02:32 AM (This post was last modified: 05/15/2009, 02:35 AM by Base-Acid Tetration.) (05/14/2009, 11:48 PM)tommy1729 Wrote: we know exp(x)-1 has a fixed point. which leads to unique half-iterate and a somewhat unique superfunction. so we might want to use the fixpoint of exp(x) - 1 for a ' surrogate fixpoint ' of exp(x). here is how - if i dont blunder - : using superfunction F(x) : F(x + 1) = exp [ F(x) ] - 1. now the simple but brilliant idea - if correct - F( x + 1 ) + 1 = exp [ F(x) ] generalize to F ( x + a ) + a = exp exp exp ... a times [ F(x) ] and Coo tetration follows !!? ok, if you add one to the argument multiple times it gets complicated: f(x+1)=exp(f(x))-1 f(x+2)=exp(exp(f(x)-1)-1= exp(e^f(x)/e)-1 ≠ exp(exp(f))-2, f(x+3)=exp(exp(exp(f(x)-1)-1)-1= e^(e^[f(x)]/e)= $\exp(\sqrt[e]{e^{f(x)}})-1 \ne \exp^{\circ 3} f(x)-3$ so f(x+1)+1 can't quite be generalized to f(x+a)+a. « Next Oldest | Next Newest »