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

  • 1
    $\begingroup$ 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

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

  • $\begingroup$ @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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.