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In other words, if we define a sequence $$ \displaystyle a_{n+1} = \sqrt{2-a_n}, \,\,\,a_0 = 0 .$$ Then, we need to find

$$ \displaystyle \prod_{n=1}^{\infty}{a_n}. $$

Well, from here I don't seem to follow. I can understand that there would be some good simplification and the product will hopefully telescope but I'm lacking the right algebra. I also thought of finding a recurrence solution probably from the corresponding DE but that didn't follow as well.

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  • $\begingroup$ have you tried to take the logarithm? $\endgroup$ – Surb Feb 16 '16 at 10:45
  • $\begingroup$ Yes, I tried. But it too didn't seem to help much. $\endgroup$ – Kartik Sharma Feb 16 '16 at 10:48
  • $\begingroup$ Have you tried first to find $(\prod a_n)^2$ ? $\endgroup$ – A. PI Feb 16 '16 at 10:52
  • $\begingroup$ Wolfram Mathematica says $\prod_{n=1}^{1000}a_n\approx 1.1547005383792515290183\ldots$ $\endgroup$ – vrugtehagel Feb 16 '16 at 11:01
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    $\begingroup$ Set $a_n=2\cos\phi_n$, then $\sqrt{2-a_n}=2\sin\frac{\phi_n}{2}=2\cos\frac{\pi-\phi_n}{2}$. Now since $\phi_{n+1}=\frac{\pi}{2}-\frac{\phi_n}{2}$, the solution for $\phi_n$ is $$\phi_n=\frac{\pi}{3}\left[1-\left(-\frac12\right)^{n+1}\right].$$ Thus even the finite product presumably telescopes. $\endgroup$ – Start wearing purple Feb 16 '16 at 11:32
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Squaring the infinite product we observe that $$ 2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots\\ = 2\cdot\frac{2}{2+\sqrt{2}}\cdot\frac{2+\sqrt{2}}{2+\sqrt{2-\sqrt{2}}} \cdot\frac{2+\sqrt{2-\sqrt{2}}}{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots \\ =\frac{4}{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} $$ But $$ \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}= 1. $$ See: Convergence of $a_{n+1}=\sqrt{2-a_n}$. Thus $$ 2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots = \frac{4}{3} $$ Finally $$ \sqrt{2}\sqrt{2-\sqrt{2}}\sqrt{2-\sqrt{2-\sqrt{2}}}\sqrt{(2-\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots = \sqrt{\frac{4}{3}} $$

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  • $\begingroup$ Where did $4$ come from? I'm not sure I get it $\endgroup$ – Yuriy S Feb 16 '16 at 11:49
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    $\begingroup$ @YuriyS All the terms in the numerator will simplify except for the first one, and the coefficient $2$. So you get $2\cdot 2 = 4$ $\endgroup$ – Kitegi Feb 16 '16 at 15:16
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Another approach is the following one: if we assume $ a_n = 2\cos(\theta_n) $ it follows that $$ \cos(\theta_{n+1})=\sqrt{\frac{1-\cos\theta_n}{2}} = \sin\left(\frac{\theta_n}{2}\right) = \cos\left(\frac{\pi-\theta_n}{2}\right)\tag{1} $$ from which we have $\theta_{n+1}=\frac{\pi-\theta_n}{2}$ and, by induction: $$ \theta_{n+k} = \frac{\pi}{3}-(-1)^k\frac{\pi}{3\cdot 2^k}+(-1)^k\frac{\theta_n}{2^k}.\tag{2}$$ Since $\theta_0=\frac{\pi}{2}$, $$ \theta_k = \frac{\pi}{3}+(-1)^k \frac{\pi}{6\cdot 2^k},\qquad \color{red}{a_k = \cos\left(\frac{\pi}{6\cdot 2^k}\right)-(-1)^k\sqrt{3}\sin\left(\frac{\pi}{6\cdot 2^k}\right)}\tag{3} $$ but since $2\cos\theta_n = \frac{\sin(2\theta_n)}{\sin(\theta_n)}$ and $\sin(\pi-\theta)=\sin(\theta)$, we also have a telescopic product.

In particular: $$ a_1\cdot a_2\cdot\ldots\cdot a_n = \frac{\sin(2\theta_1)}{\sin(\theta_n)} \tag{4}$$ hence:

$$ \prod_{n\geq 1} a_n = \frac{\sin(2\theta_1)}{\sin(\lim_{n\to +\infty}\theta_n)} = \frac{1}{\sin\frac{\pi}{3}}=\color{red}{\frac{2}{\sqrt{3}}}.\tag{5}$$

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    $\begingroup$ It's fascinating to me how closely this method parallels evaluating the Viete product, and yet how different the final result is. $\endgroup$ – Steven Stadnicki Feb 16 '16 at 17:58
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    $\begingroup$ This is a general method and that makes it really helpful. What if we didn't have a $2$ even. What if we had some $k$? $\endgroup$ – Kartik Sharma Feb 17 '16 at 8:10
  • $\begingroup$ How did you proceed from (3) to the red statement? $\endgroup$ – zz20s Feb 20 '16 at 19:02
  • $\begingroup$ @zz20s: by just computing twice the cosine of a difference through the cosine addition formulas. $\endgroup$ – Jack D'Aurizio Feb 21 '16 at 13:12
  • $\begingroup$ Ah! I see it now, thank you. $\endgroup$ – zz20s Feb 21 '16 at 21:27

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