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S Jul 22 at 4:05 history bounty ended CommunityBot
S Jul 22 at 4:05 history notice removed CommunityBot
Jul 21 at 20:08 comment added Srini I have included lot more context now. If the number of my comments here grows, I will delete my comments in this question so that the bot doesn't move it to chat. Hopefully I can manage with the comments section in my answer or the answer body itself.
Jul 21 at 18:14 answer added Malo timeline score: 3
Jul 21 at 14:48 comment added Integrand @Srini I think so, but I'm not sure. I would like to see a bit more elaboration in your answer before I award it the bounty, especially since you seem to have put a lot of thought into it.
Jul 21 at 14:46 comment added Srini If $a_n =0$, is $b_n$ always $0$? If so, I can simplify the answer I posted
Jul 20 at 21:54 answer added Srini timeline score: 3
S Jul 14 at 2:16 history bounty started Integrand
S Jul 14 at 2:16 history notice added Integrand Draw attention
Jul 13 at 16:27 history edited Integrand CC BY-SA 4.0
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Jul 13 at 16:14 history edited Integrand CC BY-SA 4.0
clarified definition
S Nov 21, 2023 at 22:07 history bounty ended CommunityBot
S Nov 21, 2023 at 22:07 history notice removed CommunityBot
Nov 14, 2023 at 3:13 history edited Alex Ravsky
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S Nov 13, 2023 at 20:49 history bounty started Integrand
S Nov 13, 2023 at 20:49 history notice added Integrand Draw attention
Dec 7, 2020 at 0:58 history edited Integrand CC BY-SA 4.0
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Dec 6, 2020 at 20:00 comment added richrow Indeed, we can simply take $f(t)=\arcsin(\sin(2\pi t))$ and since $\int_{0}^{1}f(t)dt=0$ we have $(a_n+b_n\pi)/n\to 0$. It remains to prove that $a_n,b_n\sim n$ (up to constant).
Dec 6, 2020 at 19:53 comment added richrow Probably this can be useful: sequence $(k/(2\pi))$ is equidistributed modulo 1 (en.wikipedia.org/wiki/Equidistribution_theorem) and it follows that for any Riemann-integrable function the sum $\frac{1}{N}\sum_{k=1}^{N}f(\{k/2\pi\})$ tends to the $\int_{0}^{1}f(t)dt$. I guess this can help to approximate this sum, but not sure.
Dec 6, 2020 at 18:04 history edited Integrand CC BY-SA 4.0
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Dec 6, 2020 at 13:21 answer added jjagmath timeline score: 6
Dec 6, 2020 at 7:25 answer added Claude Leibovici timeline score: 3
Dec 5, 2020 at 19:42 comment added Integrand @StevenStadnicki, I'm using $-\pi/2 \le \arcsin(x)\le \pi/2$. So for instance, $\arcsin(\sin(10))=3\pi -10$ and $\arcsin(\sin(13))=13-4\pi$. In particular, it's not as you suggest because for $n=24$ the sum vanishes but your version would give $300-72\pi$.
Dec 5, 2020 at 19:39 comment added Steven Stadnicki What range for arcsin are you using? Is this basically $\sum_k \left(k\bmod 2\pi\right)$ ?
Dec 5, 2020 at 19:34 history edited Integrand CC BY-SA 4.0
typo
Dec 5, 2020 at 17:33 comment added dezdichado have you tried restricting to the subsequence $\{b_n\neq 0\}$ and then feeding $\{b_n\}$ to the inverse sequence encyclopedia? I feel like any problem involving some sort of irrationality measure or convergents of $\pi$ either end up having a trivial recurrence or completely unsolved. Very nice question though.
Dec 5, 2020 at 17:12 history edited Integrand CC BY-SA 4.0
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Dec 5, 2020 at 17:07 history asked Integrand CC BY-SA 4.0