I know $\sum_{k=1}^{n} \sin(k)$ is bounded by a constant. How about $\sum_{k=1}^{n} \sin(k^2)$?
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$\begingroup$ you can use $sin(x) < x$ to get a bound. $\endgroup$– Srinivas KCommented Mar 27, 2015 at 17:57
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2$\begingroup$ @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. $\endgroup$– npboolCommented Mar 27, 2015 at 17:59
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2$\begingroup$ The behavior of Gauss sums seems to suggest that this is unbounded. $\endgroup$– Christian RemlingCommented Mar 27, 2015 at 18:08
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12$\begingroup$ You must be very impatient to ask the same question both here and on MSE simultaneously! math.stackexchange.com/questions/1209178/… $\endgroup$– Alex M.Commented Mar 27, 2015 at 19:42
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2$\begingroup$ @npbool : $\;\;\; \sin\hspace{-0.04 in}\Big(\hspace{-0.03 in}\frac{\pi}2\hspace{-0.03 in}\Big) = 1 \not< 1 \;\;\;\;\;\;\;\;\;$ $\endgroup$– user5810Commented Mar 27, 2015 at 21:30
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2 Answers
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No. If one selects a number $k$ at random from $1$ to a large number $n$, then for any fixed $h$, the random variables $\sin((k+1)^2), \dots, \sin((k+h)^2)$ asymptotically have mean zero, variance 1/2, and covariances 0, from standard Weyl sum estimates. Hence the variance of $\sum_{i=1}^h \sin((k+i)^2)$ is asymptotically $h/2$, which goes to infinity as $h \to \infty$. On the other hand, if the partial sums of $\sin(k^2)$ were bounded, then this variance would have to be bounded also. [Exercise: what part of the above argument breaks down when working with $\sin(k)$ instead of $\sin(k^2)$?]
It may be possible to push this argument to show that the partial sums have to fluctuate by $\gg \sqrt{n}$ infinitely often, but I haven't checked this (certainly a lower bound of $\gg n^\varepsilon$ for some small $\varepsilon>0$ should be possible from the above argument, perhaps contingent on some conjecture about the irrationality measure of $\pi$). Heuristically, the law of the iterated logarithm suggests that the sum can occasionally get as large as $\gg \sqrt{n \log\log n}$, but no larger.
answered Mar 27, 2015 at 18:26
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1$\begingroup$ Solution: $E\sin k\sin (k+1) \simeq c\not= 0$. Right, professor? $\endgroup$ Commented Mar 27, 2015 at 18:52
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1$\begingroup$ Is not it possible to find this sum explicitly as in case of the Gauss sum it looks like? $\endgroup$– SergeiCommented Apr 2, 2015 at 20:19
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4$\begingroup$ 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. $\endgroup$ Commented Apr 3, 2015 at 14:43
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$\begingroup$ 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? $\endgroup$– DigerCommented Nov 17, 2019 at 21:04
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$\begingroup$ 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 $0$. $\endgroup$ Commented Dec 22, 2020 at 9:45
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Terry Tao has already given an excellent answer to this, but I want to point out that much more is known about the partial sums of $\sin(\pi k^2 x)$ with $x$ being an irrational (such as $1/\pi$ in the question). This and related problems were studied extensively in the classical paper of Hardy and Littlewood Some Problems of Diophantine approximation. II from 1914. Hardy and Littlewood exploit the connection with the transformation formulae for $\theta$-functions, and show a number of $\Omega$ and $O$ results for such partial sums. In particular, Theorem 2.30 of their paper proves that for any irrational $x$ the series $$ \sum n^{-\alpha} \cos(n^2 \pi x), \ \text{and} \ \sum n^{-\alpha} \sin(n^2 \pi x) $$ are not convergent when $0< \alpha \le 1/2$ (and moreover are not summable by any Cesaro means). By partial summation, this implies that there are arbitrarily large values $N$ with $$ \Big| \sum_{k\le N} \sin (k^2 \pi x) \Big| \ge \frac{\sqrt{N}}{(\log N)^2}, $$ say (otherwise the series in the Hardy-Littlewood result would converge for $\alpha=1/2$), and in fact one can probably get $\gg \sqrt{N}$ from their paper (they do this explicitly for partial sums of $e^{i\pi n^2 x}$). Quadratic Weyl sums continue to be of interest -- see this very recent paper of Cellarosi and Marklof.
