try this one!

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

#MathProblem #Math

Note by Kushagraa Aggarwal
7 years, 9 months ago

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13 votes

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Comments

Let a1=12a_1 = \sqrt{\dfrac{1}{2}} and let an+1=12+12ana_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}a_n} for integers n≥1n \ge 1. We want the value of ∏n=1∞an\displaystyle\prod_{n = 1}^{\infty}a_n.

Note that a1=cos⁡(π4)a_1 = \cos \left(\dfrac{\pi}{4}\right) and if an=cos⁡(π2n+1)a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right) then an+1=12+12cos⁡(π2n+1)=cos⁡(π2n+2)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 an=cos⁡(π2n+1)a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right) for all integers n≥1n \ge 1.

Let PN=∏n=1Nan=∏n=1Ncos⁡(π2n+1)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⁡θcos⁡θ=12sin⁡2θ\sin \theta \cos \theta = \dfrac{1}{2}\sin 2\theta repeatedly, we have:

PNsin⁡(π2N+1)=sin⁡(π2N+1)∏n=1Ncos⁡(π2n+1)=12Nsin⁡(π2)=12NP_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, PN=12Ncsc⁡(π2N+1)P_N = \dfrac{1}{2^N}\csc\left(\dfrac{\pi}{2^{N+1}}\right) for all integers N≥1N \ge 1.

For x≈0x \approx 0 we have csc⁡x=1x+O(x)\csc x = \dfrac{1}{x} + O(x). Hence, PN=12N[2N+1π+O(π2N+1)]=2π+O(2−2N)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→∞N \to \infty we have 2−2N→02^{-2N} \to 0. Therefore, ∏n=1∞an=lim⁡N→∞PN=2π\displaystyle\prod_{n = 1}^{\infty}a_n = \lim_{N \to \infty}P_N = \dfrac{2}{\pi}.

Jimmy Kariznov - 7 years, 9 months ago

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thanks a lot!. it was actually not easy to get to substitute 12 \frac{1}{\sqrt{2}} by cos(π4) cos(\frac{\pi}{4})

kushagraa aggarwal - 7 years, 9 months ago

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

Thanks!

Pranav Arora - 7 years, 9 months ago

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@Pranav Arora – it was asked in a mock test.

kushagraa aggarwal - 7 years, 9 months ago

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

Pranav Arora - 7 years, 9 months ago

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@Pranav Arora – Nope , Vidyamandir Classes.

kushagraa aggarwal - 7 years, 9 months ago

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

Jeremy Shuler - 7 years, 9 months ago
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