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computing f(f(x)) = exp(x) with g(g(x)) = - exp(x)
#1
tommy1729 is back with a new idea Wink

as the title said :

computing f(f(x)) = exp(x) with g(g(x)) = - exp(x)

or at least trying , my idea is in " beta phase ".

so why should or could g(x) and f(x) be related you probably wonder.

well , as said , im not sure yet , but i will try to clarify how i got the idea.

first , the superfunction of exp(x) and - exp(x) relate.

to be more precise ; like this :

slog ( x + pi i ) = slog ( exp(x + pi i) ) - 1 = slog ( - exp (x) ) - 1

" slog ( - exp (x) ) - 1 " seems the superfunction of - exp(x).

second , - exp(x) has a real fixpoint !

thus we can use regular iteration ( or other fixpoint based methods )

third : slog(x + 2pi i) = slog(x) , thus slog(x) is periodic , we say has period 2 pi i. ( if defined at that point ! )

proof :
slog(z + 2pi i) = slog(exp( z + 2 pi i )) - 1 = slog(exp(z)) - 1 = slog(z)

fourth : slog(x) is strictly rising on the positive reals.


combining the above , we might be able to compute f(x) such that
f(f(x)) = exp(x)

or slog(x).

and maybe also arrive at the last uniqueness conditions , in the sence of a single unique slog(z) as the general solution for tetration

slog(z) + C

( C some constant )


thats the main idea , i dont expect the function to be entire however analytic continuation might be possible , although that may loose the conditions above on the domain where the continuation takes place ( but who cares ).

if analytic continuation is not possible than probably a real analytic one is still possible ... i think ...


regards

tommy1729
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computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - by tommy1729 - 06/03/2009, 09:36 PM

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