Closed form for $\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$

Question

Question: I am hoping to analytically continue the function:

$F(n) = \prod_{k=1}^{n-1} \sin(\frac{\pi k}{n})^{n-k}$

to $n=1/2$. My understanding is that means I will need a closed form for this product. What is a closed form of the above product? (Or are there other methods for performing the analytic continuation without having a closed form?)

Attempt: A similar product satisfies the identity

$\prod_{k=1}^{n-1} \sin(\frac{\pi k}{n}) = n 2^{1-n}$.

A proof of the above identity is given in another post (Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$). I have gone through the proofs given there and tried to extend the techniques used to my case, but no luck so far.

Any help is appreciated - a full solution, or just suggestions of theorems or related identities which may be of use.

Answer 1

We break up the product in a different way (analogous to changing the order of products): $$\begin{align} F(n) &= \sin{\left(\frac{\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \sin{\left(\frac{3\pi}{n}\right)} \\ &\quad \quad \vdots \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}. \end{align}$$ (Note there are $n-1$ of the first term, $n-2$ of the second, down to one of the last). Now, consider $F(n)^2$: it is sufficient to do this and square root afterwards since every term in the finite product is positive, so the whole thing must be positive. Again, we reorder the terms: $$\begin{align} F(n)^2 &= \sin{\left(\frac{(n-1)\pi}{n}\right)} \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{(n-2)\pi}{n}\right)} \dotsm \sin{\left(\frac{\pi}{n}\right)} \\ &\quad \quad \vdots \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}. \end{align}$$ (so basically I've taken the triangle of terms, flipped it over and made a rectangle). Now, $\sin{\left(\frac{(n-k)\pi}{n}\right)} = \sin{\left(\frac{k\pi}{n}\right)}$, so we can change the terms in the upper-right triangle as follows: $$\begin{align} F(n)^2 &= \sin{\left(\frac{\pi}{n}\right)} \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \times \sin{\left(\frac{2\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-1)\pi}{n}\right)} \\ &\quad \quad \vdots \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \times \sin{\left(\frac{(n-1)\pi}{n}\right)} \\ &\quad \times \sin{\left(\frac{\pi}{n}\right)} \dotsm \sin{\left(\frac{(n-2)\pi}{n}\right)} \sin{\left(\frac{(n-1)\pi}{n}\right)}. \end{align}$$ But of course this is just $n-1+1=n$ copies of $\prod_{k=1}^{n-1} \sin{\left( \frac{k\pi}{n} \right)} = n/2^{n-1}$, so we find $$ F(n) = \sqrt{\frac{n^n}{2^{n(n-1)}}} = \frac{n^{n/2}}{2^{n(n-1)/2}}, $$ as conjectured by Claude Leibovici.

Answer 2

This is not an answer since based on observation.

Looking at the values of $$P_n=\prod_{k=1}^{n-1} \sin\left(\frac{\pi k}{n}\right)^{n-k}$$ for small odd values of $n$ reveals some interesting patterns as shown in the table below $$\left( \begin{array}{cc} n & P_n \\ 3 & \frac{3 \sqrt{3}}{8} = \frac{{3^{3/2}}}{2^3}\\ 5 & \frac{25 \sqrt{5}}{1024}= \frac{{5^{5/2}}}{2^{2\times 5}} \\ 7 & \frac{343 \sqrt{7}}{2097152}= \frac{{7^{7/2}}}{2^{3\times 7}} \end{array} \right)$$ This made me thinking about $$\large{P_n=\frac{n^{n/2}}{2^{n(n-1)/2}}}$$ which seems to be verified for all $n$.

I must confess that I should be more than happy to see a proof coming here.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\prod_{k = 1}^{n - 1}\sin^{n - k}\pars{\pi k \over n} = \prod_{k = 1}^{n - 1}\sin^{n - \pars{n - k}}\pars{\pi \bracks{n - k} \over n} = \prod_{k = 1}^{n - 1}\sin^{k}\pars{\pi k \over n} \\[5mm] & = \root{\bracks{\prod_{k = 1}^{n - 1}\sin^{n - k}\pars{\pi k \over n}} \bracks{\prod_{k = 1}^{n - 1}\sin^{k}\pars{\pi k \over n}}} = \bracks{\prod_{k = 1}^{n - 1}\sin\pars{\pi k \over n}}^{n/2}\label{1}\tag{1} \end{align}

As given by the OP and in the cited link, $\ds{\prod_{k = 1}^{n - 1}\sin\pars{\pi k \over n} = {n \over 2^{n - 1}}}$. So, expression \eqref{1} becomes:

$$ \prod_{k = 1}^{n - 1}\sin^{n - k}\pars{\pi k \over n} = \pars{n \over 2^{n - 1}}^{n/2} = \bbx{\ds{n^{n/2} \over 2^{n\pars{n -1}/2}}} $$