Evaluating limit $\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}$

I stumbled across the following question which asked to evaluate...

$$\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}$$ I at first tried writing few terms $$\cos{\left(\frac {x}{2}\right)}\cos{\left(\frac {x}{4}\right)}\cos{\left(\frac {x}{8}\right)}...$$ I used the Half-angle formula to write$$\cos{\left(\frac {x}{2}\right)}=\pm\sqrt{\frac{1+\cos(x)}{2}}$$ Therefore, $$\sqrt{\frac{1+\cos(x)}{2}}\sqrt{\frac{1+\sqrt{\frac{1+\cos(x)}{2}}}{2}}...$$ As there are infinitely many two's in the denominator, the denominator goes to $\infty$ which means $$\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}=0$$

So..My question is ...Am I correct?... If not, Could you please give me some hint to how should I proceed ?

• The numerator goes to infinity too, because $\cos\left(\frac{x}{2^r}\right)$ tends to $1$. In particular, if $x=0$, the limit is $1$, not $0$. – TonyK Feb 14 '17 at 13:18
• Try looking here for some inspiration from here – Chinny84 Feb 14 '17 at 13:20
• have you tried to write $\prod=e^{\log \prod}=e^{\sum \log}$? – tired Feb 14 '17 at 13:43

Hint. One may use$$\cos{\left(\frac {x}{2^r}\right)}=\frac12 \cdot \frac{\sin{\left(\frac {x}{2^{r-1}}\right)}}{\sin{\left(\frac {x}{2^r}\right)}}$$ giving, by a telescoping product, $$\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}=\frac1{2^k}\cdot\prod_{r=1}^k\frac{\sin{\left(\frac {x}{2^{r-1}}\right)}}{\sin{\left(\frac {x}{2^r}\right)}}=\frac1{2^k}\cdot\frac{\sin{x}}{\sin{\frac {x}{2^k}}}=\frac{\large\frac{\sin x}x}{\large\frac{\sin{\frac {x}{2^k}}}{\frac {x}{2^k}}}$$ then let $k \to \infty$.
Check by induction that $$\sin(x) = 2^m\sin(2^{-m}x)\prod_{j=1}^m \cos(2^{-j} x)$$ since $$\sin(x)=2^{}\sin(2^{-1}x)\cos(2^{-1}x)=2^{2}\cos(2^{-1}x)\cos(2^{-2}x)\sin(2^{-2}x)=.......$$ Then, $$\lim_{m\to \infty } \prod_{j=1}^m \cos(2^{-j} x) = \lim_{m\to -\infty } \frac{2^{-m}x}{\sin(2^{-m}x)} \frac{\sin(x)}{x} = \frac{\sin(x)}{x}$$
• That is not the question: your $2^j$ should be $2^{-j}$. – TonyK Feb 14 '17 at 13:36