f(s),h(s) and "hairs". tommy1729 Ultimate Fellow     Posts: 1,700 Threads: 374 Joined: Feb 2009 05/20/2021, 11:54 PM " 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 tommy1729 Ultimate Fellow     Posts: 1,700 Threads: 374 Joined: Feb 2009 05/21/2021, 12:01 AM 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 MphLee Long Time Fellow    Posts: 321 Threads: 25 Joined: May 2013 05/21/2021, 05:32 PM (This post was last modified: 05/21/2021, 05:40 PM by MphLee.) 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? MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Gottfried Ultimate Fellow     Posts: 873 Threads: 128 Joined: Aug 2007 05/21/2021, 07:23 PM (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 Gottfried Helms, Kassel tommy1729 Ultimate Fellow     Posts: 1,700 Threads: 374 Joined: Feb 2009 05/22/2021, 12:24 PM 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 JmsNxn Ultimate Fellow     Posts: 977 Threads: 114 Joined: Dec 2010 05/23/2021, 12:42 AM (This post was last modified: 05/23/2021, 03:03 AM by JmsNxn.) Just a quick point. I believe this construction of hairs will only work locally. As, Is only doable for local values. This is because arctet is undefined at all periodic points, and these make up a large portion of (sort of popping up everywhere); and additionally that we can't define without a significant discussion of branch cuts (which slog, etc...). But this seems interesting. You are definitely correct though, that as we should get a flatter and flatter line; as h will get closer to looking like (if we take ). I'd just like to add that your formula should be, And for large initial values we get that for a sequence . I think iterating is necessary to construct the super logarithm in a nice way. Where if we iterate then we are equivalently iterating once we conjugate by . 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 as . That this solution is not Kneser's solution. Regards, James. « Next Oldest | Next Newest » 