- Prove that for every z the product below converges, and
\[ \cos (z / 2) \cos (z / 4) \cos (z / 8) \dots = \prod_ {k = 1} ^ {\infty} \cos (z / 2 ^ {k}) = \frac {\sin z}{z}. \]
[Hint: Use the fact that sin \(2 z = 2 \sin z \cos z . ]\)
Solution
Iterating the double-angle identity \[ \sin(2w)=2\sin w\cos w \] gives, for every positive integer \(N\), \[ \sin z =2^N\sin\!\left(\frac{z}{2^N}\right) \prod_{k=1}^N\cos\!\left(\frac{z}{2^k}\right). \] Hence, for \(z\ne0\), \[ \prod_{k=1}^N\cos\!\left(\frac{z}{2^k}\right) =\frac{\sin z/z}{\sin(z/2^N)/(z/2^N)}. \] As \(N\to\infty\), \[ \frac{\sin(z/2^N)}{z/2^N}\longrightarrow1. \] Therefore \[ \prod_{k=1}^{\infty}\cos\!\left(\frac{z}{2^k}\right) =\frac{\sin z}{z} \] for \(z\ne0\). At \(z=0\), every factor equals \(1\) and the right side is understood by its removable value \(1\). Thus the identity holds for every \(z\in\mathbb C\).