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Find : \( \sqrt{ \frac{1}{2}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} +\frac{1}{2}\sqrt{\frac{1}{2}}}}} \\ \ldots \infty \)

Note by Kushagraa Aggarwal
4 years, 3 months ago

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Let \(a_1 = \sqrt{\dfrac{1}{2}}\) and let \(a_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}a_n}\) for integers \(n \ge 1\). We want the value of \(\displaystyle\prod_{n = 1}^{\infty}a_n\).

Note that \(a_1 = \cos \left(\dfrac{\pi}{4}\right)\) and if \(a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right)\) then \(a_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}\cos\left(\dfrac{\pi}{2^{n+1}}\right)} = \cos\left(\dfrac{\pi}{2^{n+2}}\right)\).

Therefore, by induction, we have \(a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right)\) for all integers \(n \ge 1\).

Let \(P_N = \displaystyle \prod_{n = 1}^{N} a_n = \displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right)\). Using the identity \(\sin \theta \cos \theta = \dfrac{1}{2}\sin 2\theta\) repeatedly, we have:

\(P_N \sin\left(\dfrac{\pi}{2^{N+1}}\right) = \sin\left(\dfrac{\pi}{2^{N+1}}\right)\displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right) = \dfrac{1}{2^N}\sin\left(\dfrac{\pi}{2}\right) = \dfrac{1}{2^N}\).

Therefore, \(P_N = \dfrac{1}{2^N}\csc\left(\dfrac{\pi}{2^{N+1}}\right)\) for all integers \(N \ge 1\).

For \(x \approx 0\) we have \(\csc x = \dfrac{1}{x} + O(x)\). Hence, \(P_N = \dfrac{1}{2^N}\left[\dfrac{2^{N+1}}{\pi} + O\left(\dfrac{\pi}{2^{N+1}}\right)\right] = \dfrac{2}{\pi} + O(2^{-2N})\).

As \(N \to \infty\) we have \(2^{-2N} \to 0\). Therefore, \(\displaystyle\prod_{n = 1}^{\infty}a_n = \lim_{N \to \infty}P_N = \dfrac{2}{\pi}\).

Jimmy Kariznov - 4 years, 3 months ago

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thanks a lot!. it was actually not easy to get to substitute \( \frac{1}{\sqrt{2}} \) by \( cos(\frac{\pi}{4}) \)

Kushagraa Aggarwal - 4 years, 2 months ago

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If you don't mind, can you please share the source of this problem?

Thanks!

Pranav Arora - 4 years, 2 months ago

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@Pranav Arora it was asked in a mock test.

Kushagraa Aggarwal - 4 years, 2 months ago

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@Kushagraa Aggarwal Are you talking about an IIT-JEE mock test? If so, FIITJEE?

Pranav Arora - 4 years, 2 months ago

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@Pranav Arora Nope , Vidyamandir Classes.

Kushagraa Aggarwal - 4 years, 2 months ago

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Oh yeah, 2/pi, Viete's formula.

Jeremy Shuler - 4 years, 3 months ago

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