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 computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 06/03/2009, 09:36 PM tommy1729 is back with a new idea 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 « Next Oldest | Next Newest »

 Messages In This Thread computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - by tommy1729 - 06/03/2009, 09:36 PM RE: computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - by tommy1729 - 06/06/2009, 12:28 PM RE: computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - by tommy1729 - 06/06/2009, 10:28 PM

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