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exp(x) - 1
#1
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
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#2
(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)=



so f(x+1)+1 can't quite be generalized to f(x+a)+a.
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