Evaluating the infinite product $\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n})$

How does one evaluate

$\displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n})$?

Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$

$n = 1$

$\sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$

$\cos(\frac{y}{2}) = \sin(y)/2\sin(\frac{y}{2})$

$n = 2$

$\sin(\frac{y}{2})= 2\sin(\frac{y}{4})\cos(\frac{y}{4})$

$\cos(\frac{y}{4}) = \sin(\frac{y}{2})/2\sin(\frac{y}{4})$

Hence,

$\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2\sin(\frac{y}{2}) \times \sin(\frac{y}{2})/2\sin(\frac{y}{4}) ... \times \sin(\frac{y}{2^{n-1}})/2\sin(\frac{y}{2^n})$

Then some denominators/numerators cancel, reducing the expression into

$\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2^n\sin(\frac{y}{2^n})$

Is this expression no longer reducible?

Because apparently, somehow, the denominator seems to reduce to $y$

How does that work?

• – Aditya Agarwal Jul 16 '15 at 11:26
• In such questions usually need to find $\lim _{ n\rightarrow \infty }{ \prod _{ i=1 }^{ n }{ \cos { \left( \frac { x }{ 2^{ i } } \right) } } }$ which equal to $\frac { \sin { x } }{ x }$ by your method – haqnatural Jul 16 '15 at 11:30
• As $c \to \infty$, $c\sin(\frac{x}{c}) \to x$ (use L'Hospital's rule, or interpret this as a derivative by letting $u=\frac{1}{c}$, so the denominator does reduce as you say. Nice work doing the rest by the way :) – ShakesBeer Jul 16 '15 at 11:32

2 Answers

Let's take a look at the finite product first:

$$\prod_{i=0}^k \cos\frac{x}{2^i} = \frac1{2^k}\frac{\sin 2 x}{\sin \frac{x}{2^k}}$$

which is the standard trick of multiplying the product by $\sin \frac{x}{2^k}$.

Now take limit as $k \rightarrow +\infty$ and you get:

$$\prod_{i=0}^\infty \cos\frac{x}{2^i} = \frac{\sin{2 x}}{x}$$

Hint: Use the Taylor series and let $n\rightarrow \infty:$ $$2^n\sin\left(\frac{y}{2^n}\right)= 2^n\left(\frac{y}{2^n}-\left(\frac{y}{2^n}\right)^3/3! + \dots\right) = y - \frac{y^3}{3!2^{2n}} + \dots$$