$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$

Question

Please, can you help me to find the limit of the example enter image description here

where n goes to infinity ?

http://math.stackexchange.com/questions/706749/find-the-product — lab bhattacharjee, Mar 11 '14 at 09:15

score 0 · Answer 1 · answered Mar 11 '14 at 08:06

0

Hint: write cos(x/2) = sinx/(2sin(x/2)), and cos(x/4) = sin(x/2)/(2sin(x/4)), etc...

answered Mar 11 '14 at 08:06

DeepSea

77,651

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

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The answer is $\frac{\sin x}{x}$.

See to this question's #1 answer: $\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$

edited Apr 13 '17 at 12:21

Community

1

answered Mar 11 '14 at 08:11

louie mcconnell

2,126

score 0 · Answer 3 · answered Mar 11 '14 at 08:50

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Just for your information (since you received good answers for the limit), the product from $i=1$ to $i=n$ of $\cos \left(2^{-i} x\right)$ has a closed form which is $$2^{-n} \sin (x) \csc \left(2^{-n} x\right)$$

answered Mar 11 '14 at 08:50

Claude Leibovici

260,315

$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$

3 Answers3