tommy's displacement equation tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 09/16/2022, 12:22 PM When considering analytic functions and their analytic superfunctions and wanting the group addition isomo property a certain equation makes alot of sense. Consider the displacement of an infinitesimal iteration of f on the point z_0. this displacement depends on the vector displacement , or in other words ( in the context given by the first sentense ) , the derivative of f(x) minus id(x). ( By using the jacobian this idea can be extended to nonanalytic btw , see my two previous threads ) Now vector displacements naturally satistfy the group addition isomo. So we get 2 cases : Let z_0 be a parabolic fixpoint. then f(z_0) = z_0 + a z_0^2 + ... and we get the " escape equation(s) " as I like to call it : S ' (z) = f(S(z)) - z. or S ' (z + S^[-1](z_0) ) d z = f(S(z)) - z. which describes the behaviour near z_0 for infinitesimal interations h. ( this works because f(z_0) - z_0 is close to f ' (z_0) , see the pi theorem ) this is a solvable differential equation type. when z_0 is not a parabolic fixpoint and in particular not a fixpoint we must use a diffferent equation. tommy's displacement equation : For z_0 not a fixpoint and an infinitesimal complex h : f^[h](z_0) = z_0 + ( f ' (z_0) - 1 ) h  or equivalent S ' (h + S^[-1](z_0) ) dh = z_0 + ( f ' (z_0) - 1 ) h  Now the escape equation and displacement equation(s) might resemble the Julia equation but they are not the same. *** If the analytic function has no parabolic fixpoint on the complex plane and a given region A maps biholomorphic to a region B then within that range and domain the equation    S ' (h + S^[-1](z_0) ) dh = z_0 + ( f ' (z_0) - 1 ) h  Is A uniqueness condition equation... well assuming S is completely determined by its first derivate IN OTHER WORDS when it is analytic.  Then again it seems that solving the equation MUST BE ANALTYIC ? that would confirm my conjecture : tommy's group addition isomo conjecture https://math.eretrandre.org/tetrationfor...p?tid=1640 and even more it would be a strong argument for the 2sinh method ( although note the group thing is only well defined for the real iterations there ( at the moment ) ) In case of f(z) = exp this becomes  sexp ' ( s + slog(z) ) ds = z + (exp(z) - 1) s. which looks nice. but then again i see issues ... hmm regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 09/16/2022, 12:24 PM « Next Oldest | Next Newest »

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