Timeline for Finding the asymptotic rate of growth of a table of values
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 20, 2024 at 14:01 | history | bounty ended | Arbuja | ||
| S Sep 20, 2024 at 14:01 | history | notice removed | Arbuja | ||
| Sep 20, 2024 at 14:01 | vote | accept | Arbuja | ||
| Sep 19, 2024 at 16:15 | answer | added | Daniel Lichtblau | timeline score: 2 | |
| Sep 18, 2024 at 18:07 | answer | added | ydd | timeline score: 6 | |
| Sep 18, 2024 at 8:57 | comment | added | A. Kato | Three remarks on upper bound of entropy[k]. (1) U1[k] is a partition of the interval [0,1]. Therefore, entropy[k] bounded from above by the value Log[2,Length[U1[k]]], which is realized by dividing [0,1] evenly. (2) Using some knowledge of number theory, one can show that U1[k] is equal to Sum[EulerPhi[j],{j,2,k}]. This is also commented in OEIS. (3) Sum[EulerPhi[j],{j,1,k}] is known as Totient Summatory Function, and its asymptotic series is known. | |
| Sep 16, 2024 at 23:54 | comment | added | ydd | It may be helpful to consider cases where $k$ is a power of 2, and start with F[r_]:= F[r] = Range[0, r]/r and work downward with the next range table included Range[0,r-1]/r-1 etc. | |
| Sep 16, 2024 at 18:21 | history | edited | Arbuja | CC BY-SA 4.0 | Made the ending more specific. I will still accept answers which state an upperbound. |
| Sep 16, 2024 at 16:35 | comment | added | ydd | I am still having a hard time finding an approximation, but for the more basic case where F[r_]:= F[r] = Range[0, r]/r We can show that for even k, entropyEVEN[k_] := -((2^-IntegerExponent[k, 2] k Log[4] + k Log[2^(-2 + IntegerExponent[k, 2])] - 2 (-1 + k) Log[-1 + k])/( 2 (-1 + k) Log[2])) I am trying to find a way to extend this to when F[r_] := F[r] = DeleteDuplicates[Flatten[Table[Range[0, t]/t, {t, r-1, r}]]] , F[r_]:= ..., {t, r-2, r}]]], ... , F[r_]:= ..., {t, 1, r}]]] but it's quite difficult right now | |
| Sep 16, 2024 at 16:28 | comment | added | Daniel Lichtblau | I believe one can get a cheap asymptotic upper bound of 2*Log[2,k]+1. I suspect one might be able to improve on the constant term but do not know how to do that. | |
| Sep 16, 2024 at 16:07 | answer | added | azerbajdzan | timeline score: 2 | |
| Sep 16, 2024 at 16:02 | history | edited | Arbuja | CC BY-SA 4.0 | Reversed x and E using @DanielLichtblau's comment. |
| Sep 16, 2024 at 16:02 | comment | added | Arbuja | It still returns Log instead of Log2. (I'll reverse x and E.) | |
| Sep 16, 2024 at 14:50 | comment | added | Daniel Lichtblau | Ah. Maybe use Log2 instead of Log? Also I think you are reversing x and E. | |
| Sep 16, 2024 at 14:47 | comment | added | Arbuja | @DanielLichtblau The output of nlm1 return Log[x,E] instead of Log[x,2]; however, I don't know why? Perhaps, if I write nlm[x] the output will go back to Log[x,2]. | |
| Sep 16, 2024 at 14:28 | comment | added | Daniel Lichtblau | Also: How/why did Log[2,x] become Log[x,E]? I realize they are related. But what motivates that change? | |
| Sep 16, 2024 at 0:15 | answer | added | JimB | timeline score: 3 | |
| Sep 15, 2024 at 20:39 | comment | added | Daniel Lichtblau | Small comment: Because you sort by numeric values, I think the removal of duplicates is not strictly needed. Their presence will just give some extra zeros in the two lists from which P1 is formed. | |
| Sep 15, 2024 at 18:28 | history | edited | Arbuja | edited tags | |
| S Sep 15, 2024 at 18:23 | history | bounty started | Arbuja | ||
| S Sep 15, 2024 at 18:23 | history | notice added | Arbuja | Draw attention | |
| Sep 12, 2024 at 18:12 | history | asked | Arbuja | CC BY-SA 4.0 |