Timeline for Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$?
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| Dec 22, 2020 at 9:45 | comment | added | Sungjin Kim | In Christian Remling's comment, $E(\sin k \sin(k+1)) \simeq c \neq 0$ means that $$\frac1n \sum_{k\leq n} \sin k \sin (k+1) $$ $$=\frac1n \sum_{k\leq n} (-\frac12) (\cos( 2k+1) - \cos( -1))$$ $$=-\cos 1 + o(1).$$ In probability terms, this means that Covariance of the random variables is not asymptotically $0$. | |
| Nov 17, 2019 at 21:04 | comment | added | Diger | Can somebody explain to me what the issue is with $\sin(k)$ compared to $\sin(k^2)$? Both have asymptotically mean zero and variance $1/2$. The only difference is that the first case has non-vanishing covariance and the latter vanishing covariance. This would imply that $\sin(k+1),\sin(k+2),...,\sin(k+h)$ are not statistically independent. And from there? If they are not independent then the contributions can cancel to give a finite bound? | |
| Apr 3, 2015 at 14:43 | comment | added | Terry Tao | No; a incomplete Gauss sum like this one can be transformed (via Poisson summation, or the modular equation for the theta function) to what is essentially another incomplete Gauss sum, but it is only the completed Gauss sums that can be transformed into a completely explicit closed form. | |
| Apr 2, 2015 at 20:19 | comment | added | Sergei | Is not it possible to find this sum explicitly as in case of the Gauss sum it looks like? | |
| Mar 30, 2015 at 13:10 | vote | accept | npbool | ||
| Mar 27, 2015 at 18:52 | comment | added | Christian Remling | Solution: $E\sin k\sin (k+1) \simeq c\not= 0$. Right, professor? | |
| Mar 27, 2015 at 18:46 | history | edited | Terry Tao | CC BY-SA 3.0 |
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| Mar 27, 2015 at 18:26 | history | answered | Terry Tao | CC BY-SA 3.0 |