Timeline for Can iterates of a non-polynomial function be bounded by an exponential indefinitely?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2017 at 8:45 | comment | added | user78249 | By the way great answer! Never surprises me the stuff entire functions can do. | |
| Oct 28, 2017 at 8:44 | vote | accept | CommunityBot | ||
| Oct 28, 2017 at 7:52 | comment | added | user78249 | For some reason I was thinking $S(r)$ needed to be analytic. Don't know why I thought that... I do think that's a proof of Ramanujan though, that one can construct analytic functions that do this. | |
| Oct 19, 2017 at 12:28 | comment | added | js21 | any exponential --> any iterated exponential (correction) | |
| Oct 19, 2017 at 12:20 | comment | added | js21 | If $(a_n)_n$ is an increasing sequence growing faster than any exponential, then one can take a continuous piecewise linear function $g$ such that $g(a_n) = n$, and take $S(r) = r^{g(r)}$. For example $a_n = \ ^n n$ works (tetration operation). | |
| Oct 18, 2017 at 17:20 | comment | added | user78249 | Everything seems great! Took me a little while to get the punchline, but it seems straight forward now. I also am curious though how we know there is a function that tends to infinity slower than any iterated logarithm. I think I saw a proof by Ramanujan a while back (but I can't be sure if it was phrased like this, or if the result was a little different). I vaguely remember it being a proof that the iterated logarithms can't partition the growth of functions (i.e: given $f$ it isn't necessarily true that there exists $n$ with $\log^{\circ n} \le f$). Do you have a reference for this fact? | |
| Oct 18, 2017 at 15:36 | comment | added | Jean Duchon | "increasing and tends to infinity slower than any iterated logarithm": how do you know this exists? | |
| Oct 17, 2017 at 8:32 | history | answered | js21 | CC BY-SA 3.0 |