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Using empirical methods, I conjectured that$^{[1]}$$\!^{[2]}$ $$\small\prod_{k=n}^\infty\operatorname{sinc}\left({2^{-k}}\pi\right)=1-\frac{2\pi^2}9\,4^{-n}+\frac{38 \,\pi ^4}{2025}\,4^{-2n}-\frac{2332\,\pi ^6}{2679 075}\,4^{-3 n}+\frac{265618\,\pi^8 }{10247461875}\,4^{-4 n}+O\!\left(4^{-5 n}\right)\!.$$ Can we prove this? Can we find more coefficients in this expansion, or a general formula for those coefficients?

Vladimir Reshetnikov
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  • More coefficients here: https://tinyurl.com/y5shl7l6. It appears that they have alternating signs, and their absolute values decrease roughly exponentially. Powers of $\pi$ and $4$ follow the same pattern. If I'm not mistaken, the series on the right can be seen as a sparse Dirichlet series of $n$ where most of the coefficients are $0$. – Vladimir Reshetnikov May 27 '19 at 15:51
  • I was able to prove a more general result: $\prod _{k=n}^{\infty} \operatorname{sinc}\left(2^{-k} x\right)=\sum _{k=0}^{\infty} \frac{c_k,x^{2 k}}{4^{n k}},$ where $c_0=1,;c_m=\frac{1}{1-4^{-m}} \sum _{k=0}^{m-1}\frac{(-1)^{m-k} c_k}{4^k (2 m-2 k+1)!}$. I will post the details soon, unless someone beats me to it :-) I'm trying to find a non-recursive formula for $c_m$. – Vladimir Reshetnikov May 27 '19 at 22:27
  • It appears that the coefficients $c_m$ can be represented non-recursively using the double sum $\displaystyle c_m=\frac{2^{-m,(m+1)}}{\left(\frac{1}{4};\frac{1}{4}\right)m},\sum{k=0}^m,\frac{2^{-\frac{k,(3 k+1)}{2}} {m\brack k}_{1/4}}{(2 m+k+1)!},\sum _{s=0}^{2^k-1},(-1)^{t_s} (2 s+1)^{2 m+k+1},$ where ${m\brack k}_q$ is the q-binomial coefficient and $t_s$ is the Thue–Morse sequence. This might have a connection to my other question. – Vladimir Reshetnikov May 28 '19 at 01:00
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    @RodrigodeAzevedo Just a typo, it should be $n$. Corrected, thank you! – Vladimir Reshetnikov Jun 01 '19 at 18:36

1 Answers1

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Writing $\operatorname{sinc} x$ as an infinite product, taking the logarithm and changing the summation order in the triple sum gives $$\ln \prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) = \sum_{k \geq n} \ln \prod_{l \geq 1} \left( 1 - \frac {2^{-2 k}} {l^2} \right) = -\sum_{k \geq n} \sum_{l \geq 1} \sum_{m \geq 1} \frac 1 m \left( \frac {2^{-2 k}} {l^2} \right)^{\!m} = \\ \sum_{m \geq 1} c_m \frac {2^{-2 m n}} {m!}, \quad c_m = -\frac {2^{2 m} \Gamma(m) \zeta(2 m)} {2^{2 m} - 1}.$$ Then, by Faa di Bruno's formula, $$\prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) = 1 + \sum_{m \geq 1} \sum_{1 \leq l \leq m} B_{m, l}(c_1, \ldots, c_{m - l + 1}) \frac {2^{-2 m n}} {m!} = \\ 1 + \sum_{m \geq 1} B_m(c_1, c_2, \ldots, c_m) \frac {2^{-2 m n}} {m!},$$ where $B_{m, l}$ and $B_m$ are the Bell polynomials.

Maxim
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