# Tetration Forum

Full Version: f(s),h(s) and "hairs".
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" 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
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
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 eachother
I 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?
(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
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
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.