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05/26/2011, 10:09 PM
(This post was last modified: 05/26/2011, 10:21 PM by tommy1729.)
another question is : how many superfunctions can a function have ?
in this thread we have a lower and upper superfunction.
but when considering complex numbers and nonreal fixpoints and general analytic functions , i wonder about how many superfunctions one can have and how to determine them.
note that a different branch is not considered another superfunction !
also "the inverse superfunction" is quite a weird concept , since we do not know which superfunction to invert.
thinking about this seems complicated at first , but maybe i missed something trivial.
i think this question has not been asked here before and is not easy to find in books.
correct me if im wrong.
regards
tommy1729
edit : hmm , on the other hand , following the conjecture of this thread it seems the lower and upper are in fact " the same " in some sense.
so , only 1 " true superfunction " for all entire functions afterall ?
if so then the OP must be correct in the sense that the behaviour of 1 function is also seen in the other and visa versa.
then only the behaviour of 1 function needs to be shown ... in this case (thread) might have already happened.
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05/27/2011, 09:33 AM
(This post was last modified: 05/27/2011, 09:35 AM by bo198214.)
(05/26/2011, 10:09 PM)tommy1729 Wrote: another question is : how many superfunctions can a function have ?
There are different answers.
If you just ask about the number of superfunctions, then there are infinitely many. We discussed that already, when ever you have a superfunction F, F(x+1)=f(F(x)) then also the function is a superfunction, for 1periodic, this should not be new for you.
If you however ask, how many *regular* superfunctions you have at a given fixpoint, i.e. superfunction from regular fractional iterations, i.e. which have an asymptotic powerseries development at the fixpoint, which is equal to the formal fractional iteration powerseries, then there is a clear answer:
You look at the powerseries development of the corresponding function, for simplicity we assume fixpoint at 0.
, assume
Hyperbolic: : there is exactly one regular superfunction
Parabolic: : There are exactly 2(m1) regular superfunctions.
For example , that's why we have 2*(21)=2 regular superfunctions. One from left and one from right.
Generally there are petals around the fixpoint, which are alternatingly attractive and repellant (in our example coming from left is attractive and coming from right repellant), on each petal there is defined a different regular Abel function (which is the inverse of a superfunction).
The whole thing is called the LeauFatouflower and is kinda standard in holomorphic dynamics (see for example the book of Milnor mentioned on the forum).
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fascinating.
nice answer , but like most nice answers in math , it felt too short.
your answer is very enlightening , but at the same time it brings lots of questions.
holomorphic dynamics is not my thing , but i would love to learn more about it.
however it is hard to ask a formal , clear , direct and good question.
one thing that puzzles me is how the riemann surfaces of the many superfunctions relate to each other ...
also of intrest and similar to the other thread ,
if there are 2m regular superfunctions f_i from one fixpoint , is it true that at least m of those can be chosen such that
f_i(z) = f_j(z + p_q(z) + Q) ??
where p_q(z) is a periodic function and Q is a constant.
another question
if a function F(z) ( not moebius ) has only 2 fixpoints , fixpoint fp1 has N regular superfunctions and fixpoint fp2 has M regular superfunctions , then how many regular superfunctions does F(z) has at most or at minimum ?
its kinda hard to visualize all this ...
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(05/27/2011, 10:48 PM)tommy1729 Wrote: if there are 2m regular superfunctions f_i from one fixpoint , is it true that at least m of those can be chosen such that
f_i(z) = f_j(z + p_q(z) + Q) ??
where p_q(z) is a periodic function and Q is a constant.
Yes, first a (regular) superfunction is always undetermined by an xaxis shift, this is the role of Q.
Second if you have any two superfunctions f and g, then
(if g is suitably injective) is always a 1periodic function, which follows from the definition of Abel and superfunction.
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my other questions are harder i assume.
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05/28/2011, 12:25 PM
(This post was last modified: 05/28/2011, 12:26 PM by bo198214.)
(05/28/2011, 12:12 PM)tommy1729 Wrote: my other questions are harder i assume.
Well I mean in most cases the regular iteration at one fixpoint is different from the one at the other. In the case of real analytic function with two real fixpoints they are equal only in the case of fractional linear (i.e. moebius) function. This is due to a result of Karlin&McGregor, I am not aware of other "global" results about regular iteration at different fixpoints.
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could you plz provide a reference for that karlin&McGregor result ?
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05/29/2011, 09:20 PM
(This post was last modified: 05/29/2011, 09:22 PM by bo198214.)
(05/29/2011, 05:27 PM)tommy1729 Wrote: could you plz provide a reference for that karlin&McGregor result ?
Its already contained in this post.
If you also want to read some French continuing the topic, perhaps you can read:
"Etude theorique et numerique de la fonction de KarlinMcGregor", Serge Dubuc, Journal d'Analyse Math.. Vol. 42 (1982 / 83)
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05/31/2011, 07:38 PM
(This post was last modified: 05/31/2011, 07:57 PM by sheldonison.)
(05/28/2011, 09:18 AM)bo198214 Wrote: .... Second if you have any two superfunctions f and g, then
(if g is suitably injective) is always a 1periodic function, which follows from the definition of Abel and superfunction. This 1periodic function interests me. , where g(z) is an entire function, and f(z) is some other super function for the same base.
We already discussed base eta=e^(1/e), where g is the upper superfunction (cheta), and f is the regular function. Theta is a 1cyclic function defined for imag(z)>=0 as long as z isn't an integer. This is similar to the case for bases>eta, where f=sexp(z), which is tetration, and is defined from g(z), the regular superfunction, by a theta(z). Here theta(z) is the Kneser Riemann mapping function, with singularities at the integers. Theta(z) quickly decays to zero as imag(z) increases. So, theta(z) is defined if imag(z)>=0 and z is not an integer. The Schwarz reflection property is used to define sexp(z) for imag(z)<0.
For bases<eta, a different more complicated theta(z) function is involved. Take g as the upper superexponential, and f as the sexp function, with f(0)=1, f(1)=0, with b=sqrt(2). Here, once again theta(z) is not defined if z is an integer. But since f is imaginary periodic, with a period of , then there is another set of singularities for theta(z) at imag(z)=Period=~17.143I. So theta(z) is not defined outside of the area between these two lines, the real axis, and imag(z)=~17.143I. And theta(z) has singularities at integer values of z and integer values of z+Period*I.
This is much different than the standard Kneser mapping theta(z) which decays to zero as imag(z) increases. But it is possible to imagine for bases like b=sqrt(2), another theta(z) function which does decays to zero as imag(z) increases, which results in another different b=sqrt(2) superfunction with sexp like properties. See also, http://math.eretrandre.org/tetrationforu...hp?tid=515
I'm sure that theta(z) is also relevant for other superfunctions, other than exponentiation.
 Sheldon
