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+--- Thread: constant area ? (/showthread.php?tid=573)
constant area ? - tommy1729 - 01/09/2011
let U be a non-empty simply connected open subset of the complex number plane C which is not all of C , then there exists a biholomorphic mapping f from U onto the open unit disk D.
that is riemann's mapping theorem which appears to be important in the field of tetration.
now consider iterations of f.
we start with a non-empty simply connected open subset of C : U_0.
we let U_0 have area on the complex plane equal to A_0.
we map U_0 to the open unit disk U_1 by the function f.
now we define U_n = f^[n]( U_(n-1) )
and the area on the complex plane for U_n = A_n = area(D) = pi
and play with the idea of generalizing to continu iterations :
U_r = f^[r]( U_(r-1) )
area( U_r ) = A_r = area(D) = pi
clearly this is a pretty strong restriction.
i assume this has potential outside of tetration too , such as physics and calculus.
in fact , maybe this is already old hat and i forgot about it ( getting too old ? ).
it seems alot like things ive seen before ... but not exactly !
intuitively - well mine at least - it seems f^[r] will wander off in 'space' or become periodic.
by wandering off in space , i mean that lim n-> oo U_n will go to oo*.
( oo* as in oo on the riemann sphere , possible as a translation of U_x or U_x stretches into direction infinity )
what i mean by periodic is trivial ; f^[r] or equivalently U_r are periodic in r.
notice that f^[r] is weaker than f^[z] , we ( ok , I ) only require f^[r] to be real-differentiable in r.
ofcourse complex differentiable would be even nicer.
oh , before i forget , the trickiest part :
there is a bijection between U_0 and U_1.
what other bijections exist ?
when does U_x biject to U_y ?
it seems the radiuses of f^[y-x] , f^[x-y] matter alot.
further , i assume we count the overlapping area of U_a and U_b Q times , where Q is the amount of overlap.
but for that last , maybe there is a more intresting situation with a different definition.
in general a riemann mapping f and its limit n-> oo area (f^[n](U_0)) leads to 0 because of convergeance to fixpoints and oo.
but its also intresting to consider non-zero limits of limit n-> oo area (f^[n](U_0)).
another remark is that not every taylor series is an f.
this may be problematic for many approaches to this problem.
too see this consider f = exp(z).
but the logaritms of the unit circle gives a twisted riemann surface !
regards
tommy1729