Is tetration analytic? Daniel Fellow   Posts: 240 Threads: 78 Joined: Aug 2007 07/06/2022, 03:46 PM How can an analytic function grow as fast as tetration? For example the tetrates of 4,  .  Is the following identity valid?  Daniel JmsNxn Ultimate Fellow     Posts: 1,064 Threads: 121 Joined: Dec 2010 07/06/2022, 07:37 PM (07/06/2022, 03:46 PM)Daniel Wrote: How can an analytic function grow as fast as tetration? For example the tetrates of 4,  .  Is the following identity valid?  Ya, tetration is analytic... what do you mean? That's all we work on here...? That identity makes no sense, what are the coefficients, what do you mean? If they're taylor polynomials of tetration,pentation, etc... about where? And furthermore, it's highly improbable that series even converges, let alone equals the one on the left. What are you getting at? MphLee Long Time Fellow    Posts: 368 Threads: 28 Joined: May 2013 07/06/2022, 07:51 PM I'm too ignorant to even understand your second question. How the coefficients $$a_{2,k}$$ and $$b_{m,k}$$ are defined? Anyways the first lines prompts a stupid question from me... maybe full of naivety because this is out of my comfort zone. Call $$\mathcal A$$ the set of formal powerseries that converge over some open $$D$$ s.t. $$(\delta,\infty)\subseteq D$$ for some $$\delta\in \mathbb R$$. We could order it "a la Hardy" by eventual domination. Say $$f\leq g$$ iff exists an interval $$(N,\infty)$$ s.t. over that interval we have $$f(x)\le q(x)$$. Maybe here James or bo198214 can correct me: we get orders of infinity if we quotient out $$(\mathcal A, \leq)$$ by the relation $$f\sim g$$ iff $$\lim_{x\to \infty} (g-f)(x)=0$$. Now we can inject this order into the classical orders of infinity. Let $$({\mathcal C}^0(\mathbb R),\leq)$$ be the set of continuous function ordered by eventual  domination. The order of $$\mathcal A,\leq)$$ should embed, I claim, into the order of  $$({\mathcal C}^0(\mathbb R),\leq)$$. The order we are talking about is the growth rate. Question: is $$\mathcal A$$ bounded from above in $${\mathcal C}^0(\mathbb R)$$? Do exists a continuous function that bounds from above the growth rates of all the formal powerseries defined in a neighborhood of positive infinity? Or given every continuous function we can always find an analytic function growing faster? If yes... the hyperoperations are inside or outside? That is what I read when Daniel asks Quote:How can an analytic function grow as fast as tetration? MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Daniel Fellow   Posts: 240 Threads: 78 Joined: Aug 2007 07/06/2022, 08:03 PM I worked on calculations where a function grew faster than any  where k is a whole number. Greater than any but less than , my Taylor's series made sense if I could set k to . I wonder if Taylor's series are naturally more complicated than are currently used. Daniel JmsNxn Ultimate Fellow     Posts: 1,064 Threads: 121 Joined: Dec 2010 07/06/2022, 09:08 PM (This post was last modified: 07/07/2022, 12:51 AM by JmsNxn.) (07/06/2022, 07:51 PM)MphLee Wrote: I'm too ignorant to even understand your second question. How the coefficients $$a_{2,k}$$ and $$b_{m,k}$$ are defined? Anyways the first lines prompts a stupid question from me... maybe full of naivety because this is out of my comfort zone. Call $$\mathcal A$$ the set of formal powerseries that converge over some open $$D$$ s.t. $$(\delta,\infty)\subseteq D$$ for some $$\delta\in \mathbb R$$. We could order it "a la Hardy" by eventual domination. Say $$f\leq g$$ iff exists an interval $$(N,\infty)$$ s.t. over that interval we have $$f(x)\le q(x)$$. Maybe here James or bo198214 can correct me: we get orders of infinity if we quotient out $$(\mathcal A, \leq)$$ by the relation $$f\sim g$$ iff $$\lim_{x\to \infty} (g-f)(x)=0$$. Now we can inject this order into the classical orders of infinity. Let $$({\mathcal C}^0(\mathbb R),\leq)$$ be the set of continuous function ordered by eventual  domination. The order of $$\mathcal A,\leq)$$ should embed, I claim, into the order of  $$({\mathcal C}^0(\mathbb R),\leq)$$. The order we are talking about is the growth rate. Question: is $$\mathcal A$$ bounded from above in $${\mathcal C}^0(\mathbb R)$$? Do exists a continuous function that bounds from above the growth rates of all the formal powerseries defined in a neighborhood of positive infinity? Or given every continuous function we can always find an analytic function growing faster? If yes... the hyperoperations are inside or outside? That is what I read when Daniel asks Quote:How can an analytic function grow as fast as tetration? This is a problem solved by Ramanujan. It is essentially the statement that there is no slowest growing function. He works explicitly with continuous functions, and managed to construct a hierarchy of functions such that every continuous function is bounded below by one of its elements. The same proof structure can be used for fastest growing functions, there's always a bigger fish--it's just more difficult to prove with slow growing which is what makes it fascinating. The trouble starts to appear when you ask that it's analytic. Now, Ramanujan actually constructs analytic functions to solve this problem, but only refers to continuous (gotta love ramanujan). So, to answer your question, I'm going to prove that $$\mathcal{A}$$ isn't bounded rather quickly, rather than citing something. THEOREM: For every continuous function $$f$$ there exists an entire function $$g$$, such that on the real line $$f \not \ge g$$ (in the sense of your ordering). There is always an analytic function that isn't bounded above by this ordering. Proof. To begin, we can assume that $$f \neq Constant$$ in the modded out space. And therefore tends to infinity in some manner (other wise finding such $$g$$ is trivial). Secondly we can assume that $$f$$ grows faster than any polynomial, because again, $$g = \exp(x)$$ works fine. Since $$f$$ is continuous, we know that $$f(f(x))$$ is continuous and further more $$f(f(x)) \ge f(x)$$. Now define an entire function: $$g(z)\\$$ Such that: $$g(n) = f(f(n))\\$$ but otherwise $$g$$ is entire. It's always possible to construct functions like this, see the threads with Tommy about interpolation. Therefore: $$g(n) > f(n)\,\,\text{for all}\,\,n \ge N\\$$ $$\qed$$. Now, this doesn't mean that $$g$$ dominates $$f$$, it just means that $$f$$ doesn't dominate $$g$$. So to answer your question. No, $$\mathcal{A}$$ is not bounded in $$\{\mathcal{C}^0,\le\}$$. Because if this were so, pick a bound $$f$$, run this procedure to create a $$g$$ that isn't dominated by $$f$$. It gets much more difficult when you ask that it's monotone, but again, you can see a proof that there's no slowest growing monotone analytic function (Ramanujan), and simply talk about functional inverses (which preserves analycity) and we also still get that $$\mathcal{A}$$ is not bounded. In short, you can have analytic functions that grow as fast as you fucking want. MphLee Long Time Fellow    Posts: 368 Threads: 28 Joined: May 2013 07/07/2022, 10:01 PM Wooha.... sweet proof... Ramanujan, Ramanujan everywhere xDDD I forgot that thing about interpolation using Rmanujan... MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow     Posts: 1,064 Threads: 121 Joined: Dec 2010 07/08/2022, 01:31 AM (This post was last modified: 07/08/2022, 11:44 AM by JmsNxn.) (07/07/2022, 10:01 PM)MphLee Wrote: Wooha.... sweet proof... Ramanujan, Ramanujan everywhere xDDD I forgot that thing about interpolation using Rmanujan... Well you don't actually use Ramanujan to make that interpolation, you'd use a Weierstrass product. This also isn't Ramanujan's proof, his proof is more involved and actually constructs a hierarchy of slow growing functions. I just warote a quick proof to settle your question is all. But yeah, I stole a lot from ramanujan « Next Oldest | Next Newest »

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