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I am aware that there is a closed form for $\prod\limits_{k=1}^{n} \cos\left(\frac{k\pi}{n}\right)$, (see Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$).

I ran into this sum in a problem with matrix determinants with period elements, and the result is of the form $$\prod\limits_{k=1}^{n}\left(1+ W \cos\left(\frac{2k\pi}{n}\right)\right).$$

I am wondering if there is a closed form for this product, or large $n$ approximations?

I was able to find the expressions for $n=3,4,5,...,15$ in Mathematica, but I can't seem to be able to find a clear pattern.

$ \frac{1}{4} (W-2)^2,1-W,\frac{1}{16} (W (W+2)-4)^2,-\frac{1}{16} (W-1) \left(W^2-4\right)^2,\frac{1}{64} (W ((W-4) W-4)+8)^2,-\frac{1}{4} (W-1) \left(W^2-2\right)^2,\frac{1}{256} (W-2)^2 \left(W^2 (W+6)-8\right)^2,-\frac{1}{256} (W-1) \left(W^4-12 W^2+16\right)^2,\frac{((W-2) W (W ((W-4) W-20)-8)-32)^2}{1024},-\frac{1}{256} (W-1) \left(3 W^4-16 W^2+16\right)^2,\frac{((W-2) W (W+2) (W (W (W+6)-20)-8)-64)^2}{4096},-\frac{(W-1) \left(W^6-24 W^4+80 W^2-64\right)^2}{4096},\frac{\left(W^3-8 W+8\right)^2 (W (W+2) ((W-10) W+4)+16)^2}{16384},-\frac{1}{256} (W-1) \left(W^2-2\right)^2 \left(W^4-8 W^2+8\right)^2,\frac{(W (W+2) ((W-2) W (W+2) (W (W (W+6)-48)+48)-64)+256)^2}{65536},-\frac{(W-1) \left(W^2-4\right)^2 \left(W^6-36 W^4+96 W^2-64\right)^2}{65536},\frac{(W (W (W ((W-2) W (W (W ((W-8) W-56)+48)+336)-448)+1024)+256)-512)^2}{262144},-\frac{(W-1) \left(5 W^8-80 W^6+336 W^4-512 W^2+256\right)^2}{65536} $

wooohooo
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2 Answers2

6

Using Finite products of trigonometric functions with $a=1/W$, $m=1$ and $x=0$ one gets $$ \begin{align} &\prod_{k=1}^{n}\left(1+ W \cos\left(\frac{2k\pi}{n}\right)\right)\\ &\qquad = \left( \frac W2 \right)^n \cdot \prod_{k=1}^{n}\left(\frac 2W+ 2 \cos\left(\frac{2k\pi}{n}\right)\right) \\ &\qquad = \frac{W^n}{2^{n-1}} \cdot \left(T_n\left( \frac 1W \right)+(-1)^{n+1} \right) \end{align} $$ where $T_n$ is the $n$-th Chebyshev polynomial (for which various explicit expressions are known).

Example: For $n=4$ this is $$ \begin{align} & \frac{W^4}{2^3}\cdot \left(T_4\left( \frac 1W \right)-1 \right) \\ &\qquad = \frac{W^4}{8}\cdot \left(\frac{8}{W^4}-\frac{8}{W^2} \right)\\ &\qquad = 1-W^2 \end{align} $$ which coincides with $$ \begin{align} &\prod_{k=1}^{4}\left(1+ W \cos\left(\frac{k\pi}{2}\right)\right) \\ &\qquad = (1+0\cdot W)(1-1\cdot W)(1+0\cdot W)(1+1\cdot W)\\ &\qquad = 1-W^2 \, . \end{align} $$

Martin R
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Let us observe that if $T_n(x) $ is the $n$-th Chebyshev polynomial then $T_n(\cos x) =\cos nx$ and hence $x_k=\cos (2k\pi/n)$ are the roots of $T_n(x) - 1=0$.

It now follows that $y_k=1+Wx_k$ are roots of $$T_n\left(\frac{y-1}{W}\right)-1=0$$ The coefficient of $y^n$ in above equation is $2^{n-1}/W^n$ and the constant term is $T_n(-1/W)-1$. Hence the product of roots is $$(-1)^n\cdot \frac{W^n} {2^{n-1}} \cdot(T_n(-1/W)-1)$$ Since $T_n(-x) =(-1)^nT_n(x)$ the desired product is $$\frac{W^n} {2^{n-1}}\cdot(T_n(1/W)+(-1)^{n+1})$$

Paramanand Singh
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