# How to calculate this limit?lim_(n->oo)prod_(k=1)^ncos(2^(k-1)x);x!=kpi

Mar 28, 2017

$0$

#### Explanation:

$\sin \left({2}^{k} x\right) = 2 \cos \left({2}^{k - 1} x\right) \sin \left({2}^{k - 1} x\right)$

or

${f}_{k} = 2 \cos \left({2}^{k - 1} x\right) {f}_{k - 1}$
${f}_{k - 1} = 2 \cos \left({2}^{k - 2} x\right) {f}_{k - 2}$
$\cdots$
${f}_{1} = 2 \cos \left(x\right) {f}_{0}$

then

${f}_{k} = \left({2}^{k} {\prod}_{j = 0}^{k} \cos \left({2}^{j} x\right)\right) {f}_{0}$

then

${\prod}_{j = 0}^{n} \cos \left({2}^{j} x\right) = \frac{1}{2} ^ n \frac{{f}_{n}}{{f}_{0}} = \frac{1}{2} ^ n \left(\sin \frac{{2}^{n} x}{\sin} x\right)$

we can then conclude that for $x \ne k \pi$

${\lim}_{n \to \infty} {\prod}_{j = 0}^{n} \cos \left({2}^{j} x\right) = 0$

because if $x \ne k \pi$ with $k \in \mathbb{Z}$ we have

$\left\mid \sin \frac{{2}^{n} x}{\sin} x \right\mid \le M$

so there exists $n$ such that $\frac{M}{2} ^ n < \epsilon$ for arbitrarily small $\epsilon$.