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+--- Thread: f(s),h(s) and "hairs". (/showthread.php?tid=1326)
f(s),h(s) and "hairs". - tommy1729 - 05/20/2021
" Hair theory "
Let f(s) be one of those recent compositional asymtotics of tetration.
consider a = f(s_1), b = f(s_2) such that Re(s_1),Re(s_2) > 1.
Now let hair(a),hair(b) be the continu iterations paths(curves) of iterated exponentials of a,b.
So the hair/path/curves starts at a,b and then follow the direction exp^[r](a) or exp^[r](b) for real r >=0.
If f(s) was exactly tetration those paths would be flat and parallel to eachother.
So these hairs must become flatter and flatter as r grows to +oo and exp^[r](a) , exp^[r](b) = f(s_1®) , f(s_2®)
and Re(s_1®),Re(s_2®) grow also to +oo.
( because we are approximating tetration better and better )
Some questions arise :
1)
When ( if ever ) do such hairs self-intersect ?
And in particular when a,b are close to the real line ?
2)
When does hair(a) intersect with hair(b) ?
if hair(a) eventually becomes the same path (fuse*) as hair(b) or vice versa , this does not count as intersecting.
Again in particular when a,b are close to the real line ?
3) travelling a certain lenght on a hair implies how much iterations of exp ??
In the limit ( r or s going to +oo) this should probably be length 1 implies 1 iteration of exp.
In particular close to the real line.
4) A hair never splits in 2.
5) two hairs with starting points f(s_1),f(s_2) such that Re(s_1) = Re(s_2) never fuse* into the same hair.
***
Some questions are easier or harder than others. But ALL are intuitionistic imho.
These questions are both concrete and vague.
A) what superexponential is used for exp^[r] ?
B) what f(s) is used ?
But when we pick those 2 (A,B) the question is very concrete.
..and it makes sense to pick an exp^[r] based on the f(s) method.
( even though at present , even that might be a dispute or choice how to actually do it )
Yes this is very related and similar to a recent topic :
f(h(s)) = exp(f(s))
h(s) = g(exp(f(s))).
So much depends on h(s).
In fact it makes sense to define exp^[r] based on the f(s) resp h(s).
Since h(s) is close to s + 1 near the real line , the lenght argument makes sense.
And exp^[1/2](v) seems iso to ( approximate ) length increase of 1/2.
The term partition also relates here.
if an infinite set of dense hairs never intersect they locally partition a dense complex space.
regards
tommy1729
RE: f(s),h(s) and "hairs". - tommy1729 - 05/21/2021
its a bit abuse notation but hair(a) might be better described as hair(s_1) because there are many s such that f(s) = a ...
But then we need to compute the inverse f of a to find s_1 ...
Just keep it in mind.
Hair (a,s_1) might be clearer but this is also abuse notation lol
regards
tommy1729
RE: f(s),h(s) and "hairs". - MphLee - 05/21/2021
What do you mean with f? Your f in the definition satisfies f(s+1)=exp(f(s))? Or is f(s+1)=exp(f(s))+something(s)
Quote:If f(s) was exactly tetration those paths would be flat and parallel to eachotherI don't see how this can be true. If wee see hair(a) as a path (curve) I don't see it being "flat", aka a line. They are the images of rays in the complex planes (parallel to the x axis) under the function f but when f maps them I expect them to curve even in the case of tetration. Or do you mean "flat" in another way?
RE: f(s),h(s) and "hairs". - Gottfried - 05/21/2021
(05/20/2021, 11:54 PM)tommy1729 Wrote: Now let hair(a),hair(b) be the continu iterations paths(curves) of iterated exponentials of a,b.
So the hair/path/curves starts at a,b and then follow the direction exp^[r](a) or exp^[r](b) for real r >=0.
Hello Tommy, is your definition of a hair compatible with that of Devaney? I ask, because I was unable so far to completely understand what Devaney meant: a (discrete) orbit or a (continuous) trajectory... (I couldn't make it sure even by the secondary literature that I had/found so far and especially I lack the original/introductional Devaney's book)
Gottfried
RE: f(s),h(s) and "hairs". - tommy1729 - 05/22/2021
Ok sorry I was not clear enough.
A hair is a shape.
So for instance let tet(s) be analytic tetration.
a = tet(1 + i)
then consider for r > 0 : exp^[r](a).
this is a continu curve in the complex plane.
Now arctet(exp^[r](a)) = 1 + i + r.
This is a flat line with starting point 1 + i.
the SHAPE of that flat line with starting point 1 + i is a flat line starting at 1+i.
so the hair ( of exp^[r](a) with respect to tet(s) ) is this flat curve.
NOW FOR A GENERAL APPROXIMATION OF tetration f(s) we consider
hair = shape ( inverse_f(exp^[r](s_1)) ).
SO a hair is the shape (not the function and not the values of ) of h(s) where
f(h(s)) = exp(f(s)).
That is what I meant with : " If f(s) was exactly tetration those paths would be flat and parallel to eachother "
I hope this clarifies.
regards
tommy1729
RE: f(s),h(s) and "hairs". - JmsNxn - 05/23/2021
Just a quick point.
I believe this construction of hairs will only work locally. As,
\(
\text{arctet}(\exp^{[r]}(1+i)) = 1+i + r\\
\)
Is only doable for local values. This is because arctet is undefined at all periodic points, and these make up a large portion of \( \mathbb{C} \) (sort of popping up everywhere); and additionally that we can't define \( \exp^{[r]} \) without a significant discussion of branch cuts (which slog, etc...). But this seems interesting.
You are definitely correct though, that as \( \Re(s_1) \to \infty \) we should get a flatter and flatter line; as h will get closer to looking like \( s_1 + 1 + o(e^{-\lambda s_1}) \) (if we take \( f = \beta_\lambda \)). I'd just like to add that your formula should be,
\(
\text{arc}\beta_\lambda(\exp^{[r]}) = h^{[r]}(\text{arc}\beta_\lambda)
\)
And for large initial values \( s \) we get that \( h^{[r]}(s) = s+r+\sum_j o(e^{-\lambda (s_j+r)}) \) for a sequence \( s_j \).
I think iterating \( h \) is necessary to construct the super logarithm in a nice way. Where if we iterate \( h \) then we are equivalently iterating \( \exp \) once we conjugate by \( \beta_\lambda \). This is something I've been fiddling with since you posted it. I think I have to use this to describe my solution better; and once and for all show that \( \text{tet}_\beta(s) \to \infty \) as \( \Im(s) \to \infty \). That this solution is not Kneser's solution.
Regards, James.