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It is well known and somewhat easy to show that $\prod\limits_{j=1}^n \cos\frac{x}{2^j} = 2^{-n}\frac{\sin x}{\sin\frac{x}{2^n}}$.

The "2-ness" of $2$ (in $\cos\frac{x}{2^j}$) is really important to deriving this identity. I was wondering if there was anything nice known about products of the form $\prod\limits_{j=1}^n \cos\frac{x}{k^j}$ where $k$ is a positive integer $\ge 3$.

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    $\begingroup$ Would something like $\displaystyle\prod_{j = 1}^{n}(2\cos\tfrac{2x}{3^j}+1) = \dfrac{\sin x}{\sin\tfrac{x}{3^n}}$ be of interest, or do you only care about products of the form $\displaystyle\prod_{j = 1}^{n}\cos\tfrac{x}{k^j}$ for some integer $k \ge 3$? $\endgroup$ – JimmyK4542 Jan 8 '17 at 6:48
  • $\begingroup$ Unfortunately, I only care about the latter. $\endgroup$ – mathworker21 Jan 8 '17 at 7:24
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I assume this can be calculated using Chebyshev polynomials. If we use the identity $\cos\frac{x}{k^{n-j}}=T_{k^j}(\cos\frac{x}{k^n})$ and $\prod\limits_{j=1}^n \cos\frac{x}{k^j}=\prod\limits_{j=0}^{n-1} \cos\frac{x}{k^{n-j}}=\prod\limits_{j=0}^{n-1}T_{k^j}(\cos\frac{x}{k^n})$ and $2 T_m(x) T_n(x) = T_{m+n}(x) + T_{|m-n|}(x)$.

One more step: For $B=\frac{x}{k^n}, A=\cos B, T_{1}(A)*T_{k}(A)*T_{k^2}(A)*T_{k^3}(A)...=\frac{1}{2}(T_{k+1}(A) + T_{k-1}(A))*T_{k^2}(A)*T_{k^3}(A)=\frac{1}{4}(T_{k^2+(k+1)}(A)+T_{k^2-(k+1)}(A) + T_{k^2+(k-1)}(A)+T_{k^2-(k-1)}(A))*T_{k^3}(A)=\frac{1}{8}(T_{k^3+(k^2+(k+1))}(A)+T_{k^3-(k^2+(k+1))}(A)+T_{k^3+(k^2-(k+1))}(A)+T_{k^3-(k^2-(k+1))}(A) + T_{k^3+(k^2+(k-1))}(A)+T_{k^3-(k^2+(k-1))}(A)+T_{k^3+(k^2-(k-1))}(A))+T_{k^3-(k^2-(k-1))}(A))...$

For k=2 the $k^3-k^2-k-1,k^3-k^2-k+1,k^3-k^2+k-1,... $ are the list of odd numbers $1,3,5,7...,2^{n}-1$. In order to simplify the understanding lets consider the case $n=4$ $T_{1}(A)*T_{k}(A)*T_{k^2}(A)*T_{k^3}(A)=T_{1}(\cos B)+T_{3}(\cos B)+T_{5}(\cos B)+...+T_{15}(\cos B)=\cos{(1*B)} + \cos{(3*B)} + \cos{(5*B)} + ... +\cos{(15*B)}=\frac{1}{2}\frac{\sin 16B}{\sin B} $ and accordingly $\prod\limits_{j=1}^4 \cos\frac{x}{k^j}=\frac{1}{2^{4-1}}\frac{1}{2}\frac{\sin 16B}{\sin B}=\frac{1}{2^{4}}\frac{\sin x}{\sin \frac{x}{16}}$.

If we follow the prove we will see that the key point in prove was:

1) the appearing of odd numbers in the sequence of $k^3-k^2-k-1,k^3-k^2-k+1,k^3-k^2+k-1,... $ for $k=2$

2) The sum of $\cos{(1*B)} + \cos{(3*B)} + \cos{(5*B)} + ... +\cos{((2^{n}-1)*B)}=\frac{1}{2}\frac{\sin 2^n B}{\sin B}$

If we perform further investigation we will see that the point 1 can be generalized for any k>2. It will bring us to sequence defined in OEIS as: "Sequence S such that 1 is in S and if x is in S, then $k*x-1$ and $k*x+1$ are in S". See the

https://oeis.org/A147991 for k=3 $1, 2, 4, 5, 7, 11, 13, 14, 16, 20, 22, 32, 34, ....$

and https://oeis.org/A147992 for k = 4. $1, 3, 5, 11, 13, 19, 21, 43, 45, 51, 53, 75, 77, 83, 85,..$

While the generalization of point 2 is impossible. It expects the sum of cosines over this coeffs*B to bring to some definite result. But I think for k>2 this will not be a case. In order to understand my point lets consider the sum $\prod\limits_{j=1}^4 \cos\frac{x}{k^j}=2^{-3}(\cos{(43*B)} + \cos{(45*B)} + \cos{(51*B)} + ... +\cos{(85*B)})$ for $k=4$. I think there is no simpler trigonometric relation for this.

So I think for the products you expect there will not be any simple form, except the case k=2.

The exact formula for the sum is:

$\prod\limits_{j=1}^n \cos\frac{x}{k^j}=2^{-n+1}\sum\limits_{i=2^{n-1}}^{2^{n}-1}\cos{(a_{k}(i)*\frac{x}{k^n})} $

where:

for $k=2$ the $a_k(i)$ is https://oeis.org/A006257 $1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15,...$

for $k=3$ the $a_k(i)$ is https://oeis.org/A147991 $1, 2, 4, 5, 7, 11, 13, 14, 16, 20, 22, 32, 34, ...$

for $k=4$ the $a_k(i)$ is https://oeis.org/A147992 $1, 3, 5, 11, 13, 19, 21, 43, 45, 51, 53, 75, 77, 83, 85,...$

for $k=5$ the $a_k(i)$ is https://oeis.org/A153777 $1, 4, 6, 19, 21, 29, 31, 94, 96, 104, 106, 144, 146, 154, 156,...$

This $a_k(i)$ sequence can be simply generalized for any k.

And finally one amazing result: $\prod\limits_{j=0}^{n-1} \cos{k^j}=2^{-n+1}\sum\limits_{i=2^{n-1}}^{2^{n}-1}\cos{(a_{k}(i))} $

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    $\begingroup$ Can you see if you can calculate it please? It seems you mind end up with something more complicated than the original product of cosines! $\endgroup$ – mathworker21 Jan 14 '17 at 9:49
  • $\begingroup$ Actually the last cosine sum can be generalized for any k and n. If this meets your expectations. $\endgroup$ – Gevorg Hmayakyan Jan 14 '17 at 13:15
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    $\begingroup$ I just have added the general formula for sum, please take a look and if any questions will be glad to answer. $\endgroup$ – Gevorg Hmayakyan Jan 14 '17 at 22:28
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    $\begingroup$ Sure. For $k=10$ the $a_i$ is $1,9,11,89,91,109,111,889,891,909,911,1089,1091,1109,1111,8889,8891,8909,8911,9089,9091,9109,9111,10889,10891,10909,10911,11089,11091,11109,11111,88889,88891,88909,88911,89089,89091,89109,89111,90889,90891,90909,90911,91089,91091,91109,91111,108889,108891,108909,108911,109089,109091,109109,109111,110889,110891,110909,110911,111089,111091,111109,111111$ $\endgroup$ – Gevorg Hmayakyan Jan 14 '17 at 23:43
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    $\begingroup$ The mathematica command is: nxt[n_] := Flatten[10 # + {1, -1} & /@ n]; Union[ Flatten[NestList[nxt, {1}, 5]]] $\endgroup$ – Gevorg Hmayakyan Jan 14 '17 at 23:44

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