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
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?
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
