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This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $\prod_{n=1}^\infty \cos\left(\large\frac{x}{2^n}\right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that $$\large\prod_{n=1}^\infty\cos\left(\frac{x}{2^n}\right)=\frac{\sin (x)}{x}.$$

I can't figure out how to prove this.

Stackman
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Tim Raczkowski
  • 20,046

2 Answers2

22

Using the trig identity

$$\sin (2t) = 2\sin (t) \cos (t),$$

we have

$$\prod_{n = 1}^N \cos(x/2^n) = \prod_{n = 1}^N \frac{\sin(x/2^{n-1})}{2\sin(x/2^n)} = \frac{\sin(x)}{2^N\sin(x/2^N)} = \frac{\sin x}{x}\cdot \frac{x/2^N}{\sin(x/2^N)}$$

Take the limit as $N \to \infty$ and use the fact $\lim_{t\to 0} \frac{\sin t}{t} = 1$ to obtain the result.

kobe
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  • I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $\lim_{N\to\infty} 2^N \sin(2^{-N}x) = x$ – jameselmore Mar 18 '15 at 17:59
  • @jameselmore By the Taylor expansion we have $\sin(2^{-N}x)\sim_{N\to\infty}2^{-N}x$. –  Mar 18 '15 at 18:02
  • @jameselmore I've added more details. Since $(\sin t)/t \to 1$ as $t \to 0$, $(\sin x/2^N)/(x/2^N) \to 1$ as $N \to \infty$. Therefore $(x/2^N)/\sin(x/2^N) \to 1$ as $N \to \infty$. – kobe Mar 18 '15 at 18:04
  • @kobe, Thanks! Makes a lot of sense – jameselmore Mar 18 '15 at 18:06
7

Hint

$$\cos(x/2^n)=\frac12\frac{\sin(x/2^{n-1})}{\sin(x/2^n)}$$ and telescope.