Timeline for Boundedness of sum of sin(sin(n))
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5 at 17:51 | history | became hot network question | |||
| S Aug 5 at 13:15 | vote | accept | Oleksandr Liubimov | ||
| S Aug 5 at 13:15 | vote | accept | Oleksandr Liubimov | ||
| S Aug 5 at 13:15 | |||||
| Aug 5 at 13:08 | vote | accept | Oleksandr Liubimov | ||
| S Aug 5 at 13:15 | |||||
| Aug 5 at 12:34 | comment | added | mathworker21 | @AchimKrause Thanks. Don't know why I thought otherwise. | |
| Aug 5 at 11:49 | answer | added | David E Speyer | timeline score: 42 | |
| Aug 5 at 11:46 | comment | added | Achim Krause | @mathworker21 I don't think that's true: Writing $\sin(n)$ as imaginary part of $e^{in}$, the partial sums evaluate to the imaginary part of a geometric sum $e^i \cdot \frac{e^{iN}-1}{e^i-1}$, where the numerator stays bounded by $2$ no matter how large $N$ gets. | |
| Aug 5 at 11:42 | answer | added | Achim Krause | timeline score: 10 | |
| Aug 5 at 11:31 | comment | added | mathworker21 | As to your approach, I think even the partial sums of $\sin(n)$ are $\underline{\text{un}}$bounded. | |
| Aug 5 at 10:55 | history | edited | Oleksandr Liubimov | CC BY-SA 4.0 |
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| Aug 5 at 10:49 | history | edited | Oleksandr Liubimov | CC BY-SA 4.0 |
deleted 1 character in body
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| Aug 5 at 10:10 | review | Close votes | |||
| Aug 10 at 2:40 | |||||
| S Aug 5 at 9:48 | review | First questions | |||
| Aug 5 at 11:04 | |||||
| S Aug 5 at 9:48 | history | asked | Oleksandr Liubimov | CC BY-SA 4.0 |