Timeline for Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2015 at 3:50 | answer | added | Lucia | timeline score: 28 | |
| Mar 30, 2015 at 13:10 | vote | accept | npbool | ||
| Mar 27, 2015 at 21:30 | comment | added | user5810 | @npbool : $\;\;\; \sin\hspace{-0.04 in}\Big(\hspace{-0.03 in}\frac{\pi}2\hspace{-0.03 in}\Big) = 1 \not< 1 \;\;\;\;\;\;\;\;\;$ | |
| Mar 27, 2015 at 19:42 | comment | added | Alex M. | You must be very impatient to ask the same question both here and on MSE simultaneously! math.stackexchange.com/questions/1209178/… | |
| Mar 27, 2015 at 18:26 | answer | added | Terry Tao | timeline score: 65 | |
| Mar 27, 2015 at 18:08 | comment | added | Christian Remling | The behavior of Gauss sums seems to suggest that this is unbounded. | |
| Mar 27, 2015 at 18:03 | review | Low quality posts | |||
| Mar 27, 2015 at 18:05 | |||||
| Mar 27, 2015 at 17:59 | comment | added | npbool | @SrinivasK It's trivial that $sin(x)<1$. I want to know whether the partial summation is bounded by a constant, not a function. $sin(x)<x$ doesn't work here. | |
| Mar 27, 2015 at 17:57 | comment | added | Srinivas K | you can use $sin(x) < x$ to get a bound. | |
| Mar 27, 2015 at 17:46 | history | asked | npbool | CC BY-SA 3.0 |