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 ?

  • $\begingroup$ 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 $0$. $\endgroup$ – TonyK Feb 14 '17 at 13:18
  • $\begingroup$ Try looking here for some inspiration from here $\endgroup$ – Chinny84 Feb 14 '17 at 13:20
  • $\begingroup$ have you tried to write $\prod=e^{\log \prod}=e^{\sum \log}$? $\endgroup$ – 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 $k \to \infty$.

Can you take it from here?


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, $$\lim_{m\to \infty } \prod_{j=1}^m \cos(2^{-j} x) = \lim_{m\to -\infty } \frac{2^{-m}x}{\sin(2^{-m}x)} \frac{\sin(x)}{x} = \frac{\sin(x)}{x}$$

  • 2
    $\begingroup$ That is not the question: your $2^j$ should be $2^{-j}$. $\endgroup$ – TonyK Feb 14 '17 at 13:36
  • $\begingroup$ Where I don't get your point. Can you be more precise? $\endgroup$ – Guy Fsone Feb 16 '17 at 11:35
  • $\begingroup$ How strange! You edited your answer to fix the problem that I pointed out. Now you are asking me where the problem is. $\endgroup$ – TonyK Feb 16 '17 at 15:17
  • $\begingroup$ sorry I did not know that!! $\endgroup$ – Guy Fsone Feb 17 '17 at 10:57

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