exp(x) - 1

[Base-Acid Tetration]

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.