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Consider the following sum: $$ S_N = \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} \sin\left( \frac{k_1\pi}{N} \right) \sin\left( \frac{k_2\pi}{N} \right) \sin\left( \frac{k_3\pi}{N} \right) $$ The summation indexes $k_1, k_2, k_3$ are integers that satisfy the restriction $ k_1 + k_2 + k_3 =N$, and each one of them ranges between $-(N-2)$ and $N$, but they're also different from $0$.

I would be interested to know if there is any way to estimate the order of magnitude of this quantity for arbitrary large $N$ (for example, in terms of big O notation). Any reference to papers or books with similar topics would also be appreciated

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    $\begingroup$ Each term is bounded by a constant over N^3, so the sum is bounded by c/N. I have personal reason to believe c is less than 64. Gerhard "Getting Personal With Sum Estimates" Paseman, 2019.11.18. $\endgroup$ – Gerhard Paseman Nov 18 at 15:44
  • $\begingroup$ If I am not mistaken, you have a total of about 4N^2 terms, each term less than 8/N^3. They are also all positive, so I think the result is bounded by 32/N. Gerhard "Maybe It Is Even Smaller" Paseman, 2019.11.18. $\endgroup$ – Gerhard Paseman Nov 18 at 18:04
  • $\begingroup$ Also, I believe (and you should prove) that the largest term occurs when the k's are equal, giving a Max of about 17.5/N^3. Further, you should be able to bound more than half the summands by something over N^4. Finally, you can look at how many big terms there are. If there are at least N^(3/2) many, you can get a weak lower bound for your sum of N^(-3/2). Gerhard "Back Of The Envelope Rulez!" Paseman, 2019.11.18. $\endgroup$ – Gerhard Paseman Nov 18 at 18:16
  • $\begingroup$ Notice that the product of three sinuses can be replaced with just $\frac{-3}{4}\sin(\frac{2k_3\pi}N)$. $\endgroup$ – Max Alekseyev Nov 19 at 15:09
  • $\begingroup$ @Max, how does that replacement work? In particular, when the k's are nearly equal, I get a positive term. Gerhard "Read The Wrong Sign, Maybe?" Paseman, 2019.11.19. $\endgroup$ – Gerhard Paseman Nov 19 at 16:06
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I believe the problem can be approached from the generating-functions perspective.

First we notice that at least two of the $k_i$'s are positive. Based on the symmetry, let's consider the sum with $k_1,k_2>0$ and multiply it by 3. This will cover all cases, but the case with all $k_i>0$ will be counted thrice. Hence, $$S_N = 3S_N^{(2)} - 2S_N^{(1)},$$ where in $S_N^{(2)}$ sums under the assumption $k_1,k_2>0$, while $S_N^{(1)}$ sums under the assumption $k_1,k_2,k_3>0$.


Let me consider $S_N^{(1)}$, which takes the form: $$S_N^{(1)} = (\sin\alpha)^3\sum_{k_1+k_2+k_3=N\atop k_1,k_2,k_3>0} \frac{1}{k_1k_2k_3} U_{k_1-1}(\cos\alpha)U_{k_2-1}(\cos\alpha)U_{k_3-1}(\cos\alpha),$$ where $U_k()$ are Chebyshev polynomials of second kind and $\alpha:=\frac{\pi}{N}$.

From the generating function for Chebyshev polynomials we have $$\sum_{m\geq 1} \frac{U_{m-1}(\cos\alpha)}m t^m =\frac{\arctan\frac{t\sin\alpha}{1-t\cos\alpha}}{\sin\alpha}.$$ Hence, $$S_N^{(1)} = [t^N]\ (\arctan\frac{t\sin\alpha}{1-t\cos\alpha})^3.$$ So, essentially we got the generating function for $S_N^{(1)}$. It should be possible to get the asymptotic by standard means (e.g., see Section 5.4 in generatingfunctionology).


UPDATE. Here is yet another (more straightforward) take on the original problem.

Let $z:=e^{I\frac{\pi}{N}}$ and notice that $\sin(k\alpha) = -\frac{I}{2}(z^k - z^{-k})$, where $I$ is the imaginary unit. Taking into account symmetry in summation indices and that $z^N=-1$, we get \begin{split} S_N &= \frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{k_1}-z^{-k_1}) (z^{k_2}-z^{-k_2}) (z^{k_3}-z^{-k_3})\\ &=\frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} \big( (z^{k_1+k_2+k_3}-z^{-(k_1+k_2+k_3)}) - 3(z^{k_1+k_2-k_3}-z^{-(k_1+k_2-k_3)})\big)\\ &=-\frac{3I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{2k_3}-z^{-2k_3}). \end{split}

We will need generating functions: $$\sum_{k=1}^{N} \frac{t^k}{k} = \int_0^t dx \frac{1-x^N}{1-x},$$ $$\sum_{k=-(N-2)}^{-1} \frac{t^k}{k} = \sum_{k=2}^{N-1}\frac{t^{k-N}}{k-N} = \int_0^t dx\frac{1-x^{N-2}}{(1-x)x^{N-1}},$$ and sum of the two: $$F_N(t) := \sum_{-(N-2)\leq k\leq N\atop k\ne 0} \frac{t^k}{k} = \int_0^t dx \frac{1-x^{N-2}+x^{N-1}-x^{2N-1}}{(1-x)x^{N-1}} $$

Then $$S_N = -\frac{3I}{8} [t^N]\ F_N(t)^2\big( F_N(z^2t) - F_N(z^{-2}t)\big)$$

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  • $\begingroup$ @Amadocta: Your formula does not hold for $t=0$. $\endgroup$ – Max Alekseyev Nov 19 at 13:47
  • $\begingroup$ OK. Although I don't like the minus sign, I can see replacing the triple sin with a sin 2k3 term now. It seems the minus sign may be needed anyway. Gerhard "That Took Some Brain Sweat" Paseman, 2019.11.19 $\endgroup$ – Gerhard Paseman Nov 19 at 16:53
  • $\begingroup$ @GerhardPaseman: You are right being suspicious about the minus sign -- I got it wrong. Now, it's corrected. $\endgroup$ – Max Alekseyev Nov 19 at 17:03
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If you apply Gerhard's observation that you can bound each term by $C/N^3$ using $|\sin(x)|\leq|x|$ we get the following upper bound:

\begin{equation} \begin{aligned} |S_N| &\leq \sum_{k_1+k_2+k_3=N\\0<|k_i|\leq N}\frac{\pi^3}{N^3}\\ &\leq\frac{\pi^3}{N^3}\sum_{k_1+k_2+k_3=N\\0<|k_i|\leq N}1\\ &\leq\frac{\pi^3}{N^3}2N^2=\frac{2\pi^3}{N}. \end{aligned} \end{equation}

$\frac{\sin(x\pi/N)}{x}\geq 0$ for $-N\leq x\leq N$ so this should be a pretty good upper bound, at least asymptotically.

Update: I had the further idea to try to approximate the sum with an integral. Numerical tests showed that this gave a very good approximation.

\begin{equation} \begin{aligned} &\sum_{k_1+k_2+k_3=N\\|k_i|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{k_3\pi}{N})}{k_1k_2k_3}\\ &\approx \iiint_{k_1+k_2+k_3=N\\|k_i|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{k_3\pi}{N})}{k_1k_2k_3}\\ &=\iint_{|k_1|,|k_2|\leq N\\|N-k_1-k_2|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{(N-k_1-k_2)\pi}{N})}{k_1k_2(N-k_1-k_2)}\\ &=\iint_{|k_1|,|k_2|\leq N\\0<k_1+k_2\leq 2N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{(N-k_1-k_2)\pi}{N})}{k_1k_2(N-k_1-k_2)}\\ &=\left(\frac{\pi}{N}\right)\iint_{|x_1|,|x_2|\leq \pi\\0<x_1+x_2\leq 2\pi}\frac{\sin(x_1)\sin(x_2)\sin(\pi-x_1-x_2)}{x_1x_2(\pi-x_1-x_2)}\\ \end{aligned} \end{equation}

on substituting $x_1=\frac{k_1\pi}{N}$ and $x_2=\frac{k_2\pi}{N}$. The integral on the right is independent of $N$ and can be evaluated as approximately 3.589. Thus we have

\begin{equation} \begin{aligned} S_N\approx \frac{3.589 \pi}{N}=\frac{11.274}{N}. \end{aligned} \end{equation}

I have some numerical tests and this is in very close agreement with the sum for large values of N, with a relative error of less than a few % for $N$ bigger than 50 say.

Here is a plot of the approximation in Mathematika:

enter image description here

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  • $\begingroup$ I think the last line should be \begin{equation} \begin{aligned} |S_N| &\leq \frac{1}{N}\sqrt{\frac{32\log^2(N)}{N}}\\ \end{aligned} \end{equation} as otherwise it is a worse estimate than the one without oscillation $\endgroup$ – Conrad Nov 18 at 22:02
  • $\begingroup$ @Conrad Thanks for the heads up - I have updated my post to correct this and to add some numerical data. $\endgroup$ – Ivan Meir Nov 19 at 1:36
  • $\begingroup$ As Gerhard Paseman alludes to in comments on the OP, one can get a better rigorous upper bound, even more quickly, using the inequality $|(\sin y)/y| \le 1$. I recommend redoing the first part of this post. $\endgroup$ – Greg Martin Nov 19 at 6:46
  • $\begingroup$ @GregMartin Thanks I've updated my post to include this. $\endgroup$ – Ivan Meir Nov 19 at 9:41
  • $\begingroup$ I got a similar result with a slightly different change of variables $$S_N \approx \frac{1}{N} \int_{-1}^1 \int_{-x}^{2-x} \frac{\sin(x\pi)\sin(y\pi)\sin(\pi -\pi x-\pi y)}{xy(1-x-y)}dydx \approx \frac{8.53819}{N} $$ When I posted the question, I was hoping that the correct answer would turn out to be of order $O\left(\frac{1}{N^2}\right)$ (this is part of a larger problem I'm working on), but it seems pretty clear that it is not the case. $\endgroup$ – Amadocta Nov 22 at 0:58

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