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I would like to compute the infinite product $\displaystyle f(x)=\prod_{n=1}^{N\rightarrow\infty} \frac{1}{2}\left({1+\cos\frac{x}{2^n}}\right)$ for a given real $x$.

Since the terms in the product are positive and smaller than $1$, the product is bounded by $0$ and decreases. Therefore the limit $N\rightarrow\infty$ is well defined.

I know that $\displaystyle \prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$ and I wonder if $f$ also has a simple analytical expression.

Thank you.

EDIT : Using the expression $\cos^2(x)=(1+\cos(2x)/2)$ I can reducethe first sum to the second and get $f(x)=\left(\frac{\sin(x)}{x \cos(x/2)}\right)^2$

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  • $\begingroup$ Thanks for spotting the mistake, I edited. $\endgroup$ – cestyx Jun 3 '16 at 14:06
  • $\begingroup$ Wolfram says it's actually: $$\prod_{n=0}^\infty \frac{1}{2} \left( 1+\cos \frac{x}{2^n} \right)=\frac{\sin^2 x}{x^2}$$ if we start with $n=0$, not $n=1$ $\endgroup$ – Yuriy S Jun 3 '16 at 14:13
  • $\begingroup$ Yes, thanks, I solved it by reducing to the known product (I have edited with the answer) $\endgroup$ – cestyx Jun 3 '16 at 14:17
  • $\begingroup$ Well, I've already posted an answer, hope you don't mind $\endgroup$ – Yuriy S Jun 3 '16 at 14:19
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Take the well known product and square:

$$\prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$$

$$\prod_{n=1}^{\infty} \cos^2\frac{x}{2^n}=\frac{\sin^2(x)}{x^2}$$

Now use the formula:

$$\cos^2 \frac{t}{2}=\frac{1}{2} (1+\cos t)$$

To get:

$$\prod_{n=0}^\infty \frac{1}{2} \left( 1+\cos \frac{x}{2^n} \right)=\frac{\sin^2 x}{x^2}$$

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