0
$\begingroup$

I came across this question in the following form:

Compute the following infinite product

$$\left[\sin (x)\cos \left(\frac{x}{2}\right)\right]^{1/2}\cdot \left[\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right)\right]^{1/4}\cdot \left[\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right)\right]^{1/8} \cdots$$

And I have been unable to solve it or to find similar problems online, which might guide me to the solution.

However, I was able to convert it into a nested radical form, but after that I'm out of ideas. Could really use a bit of help. Thanks!

Here's the nested radical form I mentioned:

$$\sqrt{\sin (x) \cos \left(\frac{x}{2}\right) \sqrt{\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right) \sqrt {\sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right) \ldots}}}$$

$\endgroup$
1

1 Answer 1

3
$\begingroup$

Use this formula $2\sin x \cos x = \sin 2x$ recursively $$ \lim _{N \to \infty} \prod _{n=0} ^{N} \frac{S^{2^{-n-1}}_{x 2^{-n}} C^{2^{-n-1}}_{x 2^{-n-1}} 2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} }{ 2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} } = \lim _{N \to \infty} \prod _{n=0} ^{N} \frac{S^{2^{-n}}_{x 2^{-n}} } {2^{2^{-n-1}} S^{2^{-n-1}}_{x 2^{-n-1}} } $$ I believe you can start from here.

$\endgroup$
2
  • $\begingroup$ Thank you! I've arrived at the solution. $\endgroup$ May 25, 2020 at 7:50
  • $\begingroup$ really cute approach (+1)! $\endgroup$
    – G Cab
    May 25, 2020 at 8:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .