I am trying to find anything about the product $$\sin(t)\sin(2t)...\sin(nt)$$

The few things I have discovered

  • the series of this product starts from $t^n$ coefficient.
  • the first few coefficients are integers.

But I was unable to discover any details about the count and nature of those coefficients.

I need an efficient way to calculate this product. Any hint is appreciated.

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    $\begingroup$ Not the exact same but related: youtu.be/hQ532SjbxNo $\endgroup$ – Henry Lee Sep 13 '19 at 22:32
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    $\begingroup$ Yes this video opens some doors. I will update the post with my investigations based on this video. $\endgroup$ – Gevorg Hmayakyan Sep 13 '19 at 22:45
  • $\begingroup$ @Winther thanks, I think this will bring me to some solution. Now I have at least 2 ways how to calculate this. Will update the post with findings. $\endgroup$ – Gevorg Hmayakyan Sep 13 '19 at 23:35
  • $\begingroup$ I think that series is not a way to go since, by the end, you will face Pochhammer symbols as in Simply Beautiful Art's answer. $\endgroup$ – Claude Leibovici Sep 14 '19 at 5:28
  • $\begingroup$ Yes I want to get some form that will allow efficient calculation. $\endgroup$ – Gevorg Hmayakyan Sep 14 '19 at 8:26

Here is a way for finding the power-series expansion of the expression to whatever order you want, though it gets more and more tedious the higher the order (and don't see why one would bother). This doesn't give a closed form for the coefficients for arbitrary large coefficients (which would probably be some horrible long and complicated expression), but we do find how fast the coefficients will grow with $n$.

Close to $x=0$ your product can be written as

$$f_n(x) = n! x^n\exp\left(-g_n(x)\right)\,\,\text{ where }\,\, g_n(x) \equiv -\sum_{i=1}^n \log\text{sinc}(ix)$$

Using the series expansion $$\log\text{sinc}(ix)= \sum_{m\geq 1}\frac{(-1)^{m+1}4^{m}B_{2m}}{2m(2m)!}i^{2m}x^{2m}$$ where $B_{2m}$ are the Bernoulli numbers. This gives us $$g_n(x) = \sum_{m\geq 1}\frac{4^{m}|B_{2k}|}{2m(2m)!}x^{2m}S_{2m}(n)\\ = \frac{1}{6}S_2(n)x^2 + \frac{1}{180}S_4(n)x^4 + \frac{1}{2835}S_6(n) x^6 + \ldots$$ where $S_k(n) = 1 + 2^k + 3^k + \ldots + n^k$ is the sum of the first $n$ powers. From this it's not hard to derive an expression for the first few terms in the power-series of $f_n(x)$. We only need to use the first few terms in the power-series of the exponential and the first few terms in $g_n(x)$ to get the first few terms in the power-series of $f_n(x)$:

$$f_n(x) = n! x^n - \frac{n!}{6}S_2(n)x^{2+n} + n!\left[\frac{1}{72}S_2^2(n) - \frac{1}{180}S_4(n)\right]x^{4+n} \\+ n!\left[\frac{1}{1080}S_2(n)S_4(n)- \frac{1}{2835}S_6(n)-\frac{1}{6^4}S_2^3(n)\right]x^{6+n} + \ldots$$

We can use $S_2(n) = \frac{n(n+1)(2n+1)}{6}$, $S_4(n) = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ and so on to simplify this. This gives for the first few coefficients

$$[x^n] = n!$$ $$[x^{n+2}] = -n!\frac{n(n+1)(2n+1)}{36}$$ $$[x^{n+4}] = n!\frac{n(n+1)(2n+1)(50 n^3 + 39 n^2 - 11 n + 12)}{64800}$$

Note that these are not always integers. It also follows that the coefficients grows with $n$ as $[x^{n+2m}] \sim \frac{S_2^m(n)}{m!6^m}\sim \frac{n! n^{3m}}{m!18^m}$.

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  • $\begingroup$ Very interesting answer. I was unable to get anything better. Thanks. $\endgroup$ – Gevorg Hmayakyan Oct 12 '19 at 9:39
  • $\begingroup$ We can use Bell polynomials for better result. $\endgroup$ – Gevorg Hmayakyan Oct 12 '19 at 22:40

In terms of the q-Pochhammer symbol, we have


from which one can produce various identities.

As $n\to\infty$, one can show that the product tends to $0$. We can bound how fast it goes to zero by considering the following:


which gives us:


and thus,


where $\alpha=\max\{|\sin(t/2)|,|\cos(t/2)|\}$, provided that $\sin(kt)\ne0$ for any natural $k$. In the event that $\sin(t/2)$ or $\cos(t/2)$ are $1$, then the product trivially reduces down to $0$. Since the irrationality measure of $\pi$ has an upper bound of $7.6063$, we have

$$\prod_{k=1}^n|\sin(kt)|\in\mathcal O\left(n^{-3.3031}\alpha^n\right)$$

as $n\to\infty$.

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  • $\begingroup$ Thanks, for the answer,but this does not allow efficient calculation for big n-s. $\endgroup$ – Gevorg Hmayakyan Sep 14 '19 at 8:27
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    $\begingroup$ I doubt there is much that can be done concerning $t\to\infty$ as there are wild oscillations, but as $n\to\infty$ the product converges to $0$. $\endgroup$ – Simply Beautiful Art Sep 14 '19 at 21:56

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