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+--- Thread: computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) (/showthread.php?tid=302)
computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - tommy1729 - 06/03/2009
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
RE: computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - tommy1729 - 06/06/2009
i would like to note that slog cannot be entire nor meromorphic on C AND (!) denote complex iterations consistantly at the same time.
the reason is once again the period , if slog is meromorphic on C then we can solve for slog(x) = 1 , and also slog(x2) = 1 + 2pi i
and slog(x_k) = 1 + 2pi i k in general.
but exp( 1 + 2pi i k) = e for all k.
let slog(Q1) = Q2 , then complex iterations requires that for any complex Z1 :
if slog(Q1) = slog(Q1 + Z1) = Q2 , then slog(Q1 + k Z1) = Q2
assume 3 distinct Z1 Z2 Z3 and integer k1 k2 k3 :
Q2= slog(Q1) = slog(Q1 + Z1) = slog(Q1 + Z2) = slog(Q1 + Z3)
which implies :
Q2= slog(Q1 + k1 Z1) = slog(Q1 + k2 Z2) = slog(Q1 + k3 Z3 )
( also 2pi i periodic which may or may not be a Z1 Z2 Z3 value , doesnt matter )
but that is impossible !!!
since that would be a 3-periodic meromorphic function on C !
thus slog if meromorphic , must be bounded by vector additions and inequalities to avoid 3-periodic behaviour and yet be 2pi i periodic.
( since log(0) = oo entire is totally ruled out )
it might very well be that this tread has an ideal base for tetration ; for which my comments are more usefull than for other bases.
( fixed point within period or not comes to mind )
further example
assume slog is only defined for -2 pi i < im(x) < + 2 pi i
but if slog(A_k) is within this zone , and slog(A_k) = 1 + 2pi i k is solvable for all k , then this is a kind of paradox.
since then A_k has to be dense in a mainly vertical way.
which loses the property of Coo ( unless lineair ) at that set !!!
in fact , if A_k are ' close ' to eachother , it tends to be constant in that zone.
so we end up with - i guess -
slog(z) is defined for -2 pi i < im(z) < + 2 pi i
and maps only to -2 pi i < im(slog(z)) < + 2 pi i
regards
tommy1729
RE: computing f(f(x)) = exp(x) with g(g(x)) = - exp(x) - tommy1729 - 06/06/2009
(06/06/2009, 12:28 PM)tommy1729 Wrote: i would like to note that slog cannot be entire nor meromorphic on C AND (!) denote complex iterations consistantly at the same time.
the reason is once again the period , if slog is meromorphic on C then we can solve for slog(x) = 1 , and also slog(x2) = 1 + 2pi i
and slog(x_k) = 1 + 2pi i k in general.
but exp( 1 + 2pi i k) = e for all k.
let slog(Q1) = Q2 , then complex iterations requires that for any complex Z1 :
if slog(Q1) = slog(Q1 + Z1) = Q2 , then slog(Q1 + k Z1) = Q2
assume 3 distinct Z1 Z2 Z3 and integer k1 k2 k3 :
Q2= slog(Q1) = slog(Q1 + Z1) = slog(Q1 + Z2) = slog(Q1 + Z3)
which implies :
Q2= slog(Q1 + k1 Z1) = slog(Q1 + k2 Z2) = slog(Q1 + k3 Z3 )
( also 2pi i periodic which may or may not be a Z1 Z2 Z3 value , doesnt matter )
but that is impossible !!!
since that would be a 3-periodic meromorphic function on C !
thus slog if meromorphic , must be bounded by vector additions and inequalities to avoid 3-periodic behaviour and yet be 2pi i periodic.
( since log(0) = oo entire is totally ruled out )
it might very well be that this tread has an ideal base for tetration ; for which my comments are more usefull than for other bases.
( fixed point within period or not comes to mind )
further example
assume slog is only defined for -2 pi i < im(x) < + 2 pi i
but if slog(A_k) is within this zone , and slog(A_k) = 1 + 2pi i k is solvable for all k , then this is a kind of paradox.
since then A_k has to be dense in a mainly vertical way.
which loses the property of Coo ( unless lineair ) at that set !!!
in fact , if A_k are ' close ' to eachother , it tends to be constant in that zone.
so we end up with - i guess -
slog(z) is defined for -2 pi i < im(z) < + 2 pi i
and maps only to -2 pi i < im(slog(z)) < + 2 pi i
regards
tommy1729
seems i wrote slog(z) , i meant inv slog(z) sorry.
not used to this 'new function' yet
riemann surfaces should be handy i think ...
still thinking ...
tommy1729