I would like to compute the infinite product $\displaystyle f(x)=\prod_{n=1}^{N\rightarrow\infty} \frac{1}{2}\left({1+\cos\frac{x}{2^n}}\right)$ for a given real $x$.
Since the terms in the product are positive and smaller than $1$, the product is bounded by $0$ and decreases. Therefore the limit $N\rightarrow\infty$ is well defined.
I know that $\displaystyle \prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$ and I wonder if $f$ also has a simple analytical expression.
Thank you.
EDIT : Using the expression $\cos^2(x)=(1+\cos(2x)/2)$ I can reducethe first sum to the second and get $f(x)=\left(\frac{\sin(x)}{x \cos(x/2)}\right)^2$