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

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