Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$?

Question

I recently came upon the following expression in an attempt at getting a closed-form solution for a recursive relation: $$\prod_{i=1}^n \cos(a^i\theta)$$ where $a<1$. Is there a way to make this product into a sum or otherwise make it simpler, or approximate it? In particular for the problem I was looking at, $a$ was $\frac{3}{4}$ and $\theta$ was $\frac{\pi}{4}$. Thanks!

Answer 1

I haven't worked this out in full detail, but I think it will do what you need. Start by taking the logarithm. With $$f(\theta)=\prod_{k=1}^n\cos(a^k\theta),$$ we have $$\log f(\theta) = \sum_{k=1}^n\log\cos(a^k\theta)=-\sum_{k=1}^n\int_0^{a^k\theta}\tan x \mathrm{dx}$$ Now integrate the Maclaurlin series$$ \tan x = x + \frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\dots$$ to approximate the sum. This is easy, since we just have sums of geometric series. Finally, take the exponential of the sum to compute $f(\theta).$

For numerical computation, I doubt this would be any better than simply evaluating the product directly, but for analyzing the behavior of $f$ I think it will work out, once one slogs through all the details. You can, of course, get precise bounds by using the remainder term in the Maclaurin series.

Answer 2

from $$ \cos z = {{e^{\,i\,z} + e^{\, - \,i\,z} } \over 2} = {1 \over 2}e^{\,i\,z} \left( {1 + e^{\, - \,i\,2z} } \right) $$ we get $$ \eqalign{ & \ln \cos z = \ln {1 \over 2} + i\,z + \ln \left( {1 + e^{\, - \,i\,2z} } \right) = \cr & = \ln {1 \over 2} + i\,z + \ln \left( {1 + 1 - i2z - 2z^2 + O\left( {z^3 } \right)} \right) = \cr & = \ln {1 \over 2} + i\,z + \ln \left( {2\left( {1 - iz - z^2 + O\left( {z^3 } \right)} \right)} \right) = \cr & = \ln {1 \over 2} + i\,z - \ln 2 - iz - {{z^2 } \over 2} + O\left( {z^3 } \right) = \cr & = - {{z^2 } \over 2} + O\left( {z^3 } \right) \cr} $$

Therefore we can say that $$ \eqalign{ & \ln P(x,a,n) = \ln \prod\limits_{k = 1}^n {\cos (a^{\,k} x)} = \sum\limits_{k = 1}^n {\ln \cos (a^{\,k} x)} = \cr & = - {1 \over 2}\sum\limits_{k = 1}^n {\left( {x^{\,2} a^{\,2k} + O\left( {a^{\,3k} x^3 } \right)} \right)} = - {{x^{\,2} a^{\,2} } \over 2}{{1 - a^{2n} } \over {1 - a^{\,2} }} + O\left( {a^{\,3} x^3 } \right) \cr} $$

Answer 3

Note this is @saulspatz answer with the computations added (I couldn't resist). With $$ f(\theta)=\prod_{k=1}^n\cos(a^k\theta) $$ Then noting $$ -\int_0^{a^k\theta}\tan x\, \mathrm{dx}=\log\cos(a^k\theta) $$ we have on taking the logarithm \begin{align*} \log f(\theta) &= \sum_{k=1}^n\log\cos(a^k\theta)=\log\cos(a\theta)+\dotsb+\log\cos(a^n\theta)\\ &=-\sum_{k=1}^n\int_0^{a^k\theta}\tan x \,\mathrm{dx}\\ &=-\sum_{k=1}^n\int_0^{a^k\theta}x + \frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\dots\, \mathrm{dx}\quad\text{(by the Maclaurlin series for $\tan{x}$)}\\ &=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta} +\dots\right)\\ \end{align*} Note each of these sums are all finite geometric sums: For the first \begin{align*} \sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta} &=\frac{\theta^2}{2}(a^2+a^4+\dotsb+a^{2n})\\ &=\frac{a^2\theta^2}{2}(1+a^2+\dotsb+a^{2(n-1)})\\ &= \frac{a^2\theta^2}{2}\cdot\frac{1-a^{2n}}{1-a^2}=\frac{a^2\theta^2}{2}\cdot\frac{(1-a^{n})(1+a^{n})}{(1-a)(1+a)}\\ \end{align*} Hence \begin{align*} \log f(\theta) &=-\left(\sum_{k=1}^n\Big{[}\frac{x^2}{2}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{x^4}{12}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{x^6}{45}\Big{]}_0^{a^k\theta} +\sum_{k=1}^n\Big{[}\frac{17x^8}{2520}\Big{]}_0^{a^k\theta} +\dots\right)\\ &=\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}+\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}+\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}+\frac{17a^8\theta^8}{2520}\cdot\frac{a^{8n}-1}{1-a^8}+\dotsb \end{align*} Now take the exponential of the sum to compute $f(\theta)$: \begin{align*} f(\theta) &= \exp\left(\frac{a^2\theta^2}{2}\cdot\frac{a^{2n}-1}{1-a^2}\right)\cdot \exp\left(\frac{a^4\theta^4}{12}\cdot\frac{a^{4n}-1}{1-a^4}\right)\cdot \exp\left(\frac{a^6\theta^6}{45}\cdot\frac{a^{6n}-1}{1-a^6}\right)\\ &\qquad\qquad\qquad\qquad\cdot\exp\left(\frac{17a^8\theta^8}{2520} \cdot\frac{a^{8n}-1}{1-a^8}\right)\dotsb\\ &=\prod_{k=1}^{\infty}\exp\left(\frac{\tan^{(2k-1)}(0)\,a^{2k}\theta^{2k}}{2k(2k-1)!}\cdot\frac{a^{2kn}-1}{1-a^{2k}}\right) \end{align*} where $\tan^{(2k-1)}(0)$ is the $(2k-1)$th derivative of $\tan x$ evaluated at $0$.