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Let $n$ be a positive integer. Let $b = 2 n - 1$. Let $x$ be a positive integer.

Define $f(x)$ as :

$$ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $$

Then it appears that

$$ f(b) = \frac{4}{5} + C(n)$$

And $C(n)$ is close to zero.

In fact

$$ \lim_{n \to \infty} C(n) = 0 $$

How do we prove this ??

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  • $\begingroup$ It is known that $ \int ln( sin(x) + 5/4 ) = 0 $ over its period $2 \pi$ so I guess that hints at it , however that is no proof right ? $\endgroup$ – mick Sep 20 at 17:03
  • $\begingroup$ Related ??? math.stackexchange.com/questions/2075374/… $\endgroup$ – mick Sep 20 at 17:06
  • $\begingroup$ ofcourse the law of large numbers suggests this. So "statistically" it must be true. $\endgroup$ – mick Sep 20 at 18:10
  • $\begingroup$ it would indicate that the $n^\text{th}$ root of the product tends to $1$, but that result is not precise enough to get the answer sought. $\endgroup$ – robjohn Sep 21 at 18:36
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Using $$ \begin{align} \sin\left(\frac{k\pi}n\right)+\frac54 &=\frac{e^{ik\pi/n}-e^{-ik\pi/n}}{2i}+\frac54\tag1\\ &=\frac{e^{-ik\pi/n}}{2i}\left(e^{ik\pi/n}+2i\right)\left(e^{ik\pi/n}+i/2\right)\tag2 \end{align} $$ we get $$ \begin{align} \prod_{k=1}^{2n-1}\left(\sin\left(\frac{k\pi}n\right)+\frac54\right) &=\frac45\prod_{k=0}^{2n-1}\color{#00F}{\frac{e^{-ik\pi/n}}{2i}}\color{#C00}{\left(e^{ik\pi/n}+2i\right)}\color{#090}{\left(e^{ik\pi/n}+i/2\right)}\tag3\\ &=\frac45\color{#00F}{\frac{(-1)^{n-1}}{2^{2n}}}\color{#C00}{\left[z^{2n}-1\right]_{z=2i}}\color{#090}{\left[z^{2n}-1\right]_{z=i/2}}\tag4\\[3pt] &=\frac45\frac{(-1)^{n-1}}{2^{2n}}\left(2-(-1)^n\left(2^{2n}+2^{-2n}\right)\right)\tag5\\[6pt] &=\frac45+\color{#90F}{\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)}\tag6\\[9pt] &=\frac45+\color{#90F}{C(n)}\tag7 \end{align} $$ Explanation:
$(3)$: $\frac45$ of the $k=0$ term is $1$; apply $(2)$
$(4)$: $\prod\limits_{k=0}^{2n-1}\!\left(z+e^{ik\pi/n}\right)=z^{2n}-1$
$(5)$: evaluate at $z=2i$ and $z=i/2$
$(6)$: simplify
$(7)$: the product is $\frac45+C(n)$

Thus, $$ \begin{align} C(n) &=\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)\tag8\\[6pt] &=O\!\left(4^{-n}\right)\tag9 \end{align} $$

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  • 1
    $\begingroup$ Note that $(6)$ and $(8)$ are exact closed forms. $\endgroup$ – robjohn Sep 21 at 18:27
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If you are allowed to use the complex representation of the sine and Pochhammer symbols, $$f(b) = \prod_{i = 1}^{2n-1} \Big[\sin \left(\frac{\pi i}{n}\right)+\frac{5}{4}\Big]=-\frac{4}{5} \frac{ \left(\frac{i}{2};e^{\frac{i \pi }{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi n}}$$which makes $$C_n=-\frac{4}{5} \Big[1+\frac{ \left(\frac{i}{2};e^{\frac{i \pi }{n}}\right){}_{2 n} \left(2 i;e^{\frac{i \pi }{n}}\right){}_{2 n}}{4^n\,e^{i \pi n}}\Big]$$ For the angles we know the exact trignometric functions, this generates for $C_n$'s the sequence $$\left\{\frac{9}{20},-\frac{31}{320},\frac{129}{5120},-\frac{511}{81920},?,-\frac{8191}{20971520},?,-\frac{131071}{5368709120},?,?,?,-\frac{33554431}{351843720888320}\right\}$$ and $$\log(|C_n|) \sim \frac 12 - b\,n \qquad \text{where} \qquad b=1.38861 \quad \left(\sigma_b=0.00053\right)$$

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  • $\begingroup$ (+1) I compared $C(n)=\frac{(-1)^{n-1}2^{3-2n}+2^{2-4n}}5$ and the values you have match. $\endgroup$ – robjohn Sep 21 at 17:13
  • $\begingroup$ $\log(8/5)=0.47000363$ which is close to $\frac12$. $\log(4)=1.38629436$ which is close to $1.38861$. $\endgroup$ – robjohn Sep 21 at 17:19

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