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+--- Thread: exp(x) - 1 (/showthread.php?tid=293)
exp(x) - 1 - tommy1729 - 05/14/2009
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
RE: exp(x) - 1 - Base-Acid Tetration - 05/15/2009
(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.