Explained: "Task is to find∏k=0∞cos⁡(x⋅2−k)"

amantantawq5l

Answered question

2022-04-03

Task is to find

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

zalutaloj9a0f

Beginner2022-04-04Added 17 answers

For the partial products we have

\prod_{k = 0}^{N} \cos \frac{x}{2^{k}} = \frac{\sin \frac{x}{2^{N}} \cos \frac{x}{2^{N}}}{\sin \frac{x}{2^{N}}} \prod_{k = 0}^{N - 1} \cos \frac{x}{2^{k}}

= \frac{\sin \frac{x}{2^{N - 1}}}{2 \sin \frac{x}{2^{N}}} \prod_{k = 0}^{N - 1} \cos \frac{x}{2^{k}}

= \frac{\sin \frac{x}{2^{N - 1}} \cos \frac{x}{2^{N - 1}}}{2 \sin \frac{x}{2^{N}}} \prod_{k = 0}^{N - 2} \cos \frac{x}{2^{k}}

= \frac{\sin \frac{x}{2^{N - N}}}{2^{N} \sin \frac{x}{2^{N}}} \cos \frac{x}{2^{0}}

= \frac{\sin x \cos x}{2^{N} \sin (x \cdot 2^{- N})}

From that, the limit is easily found.

membatas0v2v

Beginner2022-04-05Added 19 answers

Using the identity

\cos (x 2^{- k}) = \frac{\sin (x 2^{1 - k})}{2 \sin (x 2^{- k})}

we get

\prod_{k = 0}^{n} \cos (x 2^{- k}) = \prod_{k = 0}^{n} \frac{\sin (x 2^{1 - k})}{2 \sin (x 2^{- k})}

= \frac{1}{2^{n + 1}} \frac{\sin (2 x)}{\sin (x 2^{- n})}

Now use the limit

lim_{x \to 0} \frac{\sin (x)}{x} = 1

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?