Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$? 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!
 A: 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.
A: 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
$$
\eqalign{
  & \ln P(x,a,n) = \ln \prod\limits_{k = 1}^n {\cos (a^{\,k} x)}  = \sum\limits_{k = 1}^n {\ln \cos (a^{\,k} x)}  =   \cr 
  &  =  - {1 \over 2}\sum\limits_{k = 1}^n {\left( {x^{\,2} a^{\,2k}  + O\left( {a^{\,3k} x^3 } \right)} \right)}  =  - {{x^{\,2} a^{\,2} } \over 2}{{1 - a^{2n} } \over {1 - a^{\,2} }} + O\left( {a^{\,3} x^3 } \right) \cr} 
$$
A: 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 $0$.
