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+--- Thread: tommy's displacement equation (/showthread.php?tid=1641)
tommy's displacement equation - tommy1729 - 09/16/2022
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/tetrationforum/showthread.php?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
RE: tommy's displacement equation - tommy1729 - 09/16/2022
related :
https://math.eretrandre.org/tetrationforum/showthread.php?tid=1639