Double Trouble

Calculus Level 5

S n = ( 1 − tan ⁡ 4 π 2 3 ) ( 1 − tan ⁡ 4 π 2 4 ) ( 1 − tan ⁡ 4 π 2 5 ) … ( 1 − tan ⁡ 4 π 2 n ) S_n = \left(1-\tan^4\frac{\pi}{2^3}\right) \left(1-\tan^4\frac{\pi}{2^4}\right) \left(1-\tan^4\frac{\pi}{2^5}\right) \ldots \left(1-\tan^4\frac{\pi}{2^n}\right)

For S n S_n as defined above, evaluate 30.28 lim ⁡ n → ∞ π 3 S n \displaystyle30.28 \lim_{n\to\infty}\frac{\pi^3}{S_n} to the nearest Integer .


The answer is 969.

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Anish Puthuraya by Anish Puthuraya , 24, India

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1 solution

Alakh Aggarwal
May 13, 2014

taking one term out of bracket.... S(n) = (1-tan^4 pi/2^3)(1-tan^4 pi/2^4)..............(1-tan^4 pi/2)n) series has n-2 terms......... analysing (k-2)th term.......... 1- tan^4 pi/2^k = (1 - tan^2 pi/2^k)(1 + tan^2 pi/2^k) = {(cos^2 pi/2^k - sin^2 pi/2^k)/cos^2 pi/2^k)}(sec^2 pi/2^k) = cos(pi / 2^(k-1) )/cos^4 pi/2^k therefore, series --> S(n) = cos (pi/4) cos(pi/8)........cos(pi/2^(n-1) / { cos(pi/8)cos(pi/16).....cos(pi/2^n) }^4 = sin(2^(n-2)pi/2^(n-1)) / 2^(n-2)sin(pi/2^(n-1)) * {2^(n-2)sin(pi/2^(n))/sin(2^(n-2)pi/2^(n))}^4 [using cos x . cos 2x . cos 4x. ................ cos 2^n x = sin(2^n x)/2^n sinx ].............. now putting lim n--> infinity and using lim x--> 0 sinx / x=1 S(n) = 2 pi^3 / 64 = pi^3 / 32 therefore, 30.28 lim(n--> infinity) pi^3 / S(n) = 30.28 * 32 = 968.96

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I also did the same , but why I am getting the answer different?

S n = ( cos ⁡ 4 π 2 3 − sin ⁡ 4 π 2 3 cos ⁡ 4 π 2 3 ) ( cos ⁡ 4 π 2 4 − sin ⁡ 4 π 2 4 cos ⁡ 4 π 2 4 ) … ( cos ⁡ 4 π 2 n − sin ⁡ 4 π 2 n sin ⁡ 4 π 2 n ) \displaystyle S_n = \left(\dfrac{\cos^4\frac{\pi}{2^3}-\sin^4\frac{\pi}{2^3}}{\cos^4\frac{\pi}{2^3}}\right)\left(\dfrac{\cos^4\frac{\pi}{2^4} -\sin^4\frac{\pi}{2^4}}{\cos^4\frac{\pi}{2^4}}\right)\ldots\left(\dfrac{\cos^4\frac{\pi}{2^n}-\sin^4\frac{\pi}{2^n}}{\sin^4\frac{\pi}{2^n}}\right)

S n = ( cos ⁡ 2 π 2 3 − sin ⁡ 2 π 2 3 cos ⁡ 4 π 2 3 ) ( cos ⁡ 2 π 2 4 − sin ⁡ 2 π 2 4 cos ⁡ 4 π 2 4 ) … ( cos ⁡ 2 π 2 n − sin ⁡ 2 π 2 n sin ⁡ 4 π 2 n ) \displaystyle S_n = \left(\dfrac{\cos^2\frac{\pi}{2^3}-\sin^2\frac{\pi}{2^3}}{\cos^4\frac{\pi}{2^3}}\right)\left(\dfrac{\cos^2\frac{\pi}{2^4} -\sin^2\frac{\pi}{2^4}}{\cos^4\frac{\pi}{2^4}}\right)\ldots\left(\dfrac{\cos^2\frac{\pi}{2^n}-\sin^2\frac{\pi}{2^n}}{\sin^4\frac{\pi}{2^n}}\right)

S n = cos ⁡ π 2 2 × cos ⁡ π 2 3 ⋯ × cos ⁡ π 2 n − 1 ( cos ⁡ π 2 3 × π 2 4 ⋯ cos ⁡ π 2 n ) \displaystyle S_{n} = \dfrac{ \cos \dfrac{\pi}{2^2} \times \cos \dfrac{\pi}{2^3} \cdots \times \cos \dfrac{\pi}{2^{n-1}}}{\left(\cos \dfrac{\pi}{2^3} \times \dfrac{\pi}{2^4} \cdots \cos \dfrac{\pi}{2^n}\right)}

S n = sin ⁡ π 2 2 n − 1 sin ⁡ π 2 n − 1 ( sin ⁡ π 4 2 n sin ⁡ π 2 n ) 4 \displaystyle S_{n} = \dfrac{ \dfrac{\sin \dfrac{\pi}{2}}{2^{n-1} \sin \dfrac{\pi}{2^{n-1}}}}{\left( \dfrac{ \sin \dfrac{\pi}{4}}{2^n \sin \dfrac{\pi}{2^n}}\right)^4}

lim ⁡ → ∞ S n = 1 π 1 ( 2 π ) 4 \displaystyle \lim_{\to \infty } S_{n} = \dfrac{ \dfrac{1}{\pi}}{\dfrac{1}{(\sqrt{2} \pi)^4}}

S n = 4 ( π ) 3 S_{n} = 4 (\pi)^3

30.28 π 3 4 π 3 = 30.28 × 4 = 121.12 w h i c h i s n e a r l y e q u a l t o 121 30.28 \dfrac{\pi^3}{4 \pi^3} = 30.28 \times 4 = 121.12 ~which~is~nearly~equal~to~ 121

Calvin Lin Anish Puthuraya

U Z - 6 years, 4 months ago

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