Australian Mathematical Olympiad 2010

Algebra Level 4

sin ⁡ ( x n + 1 − x n ) + 3 1 − 2 n 2 sin ⁡ ( x n ) sin ⁡ ( x n + 1 ) = 0 \large\ \sin {( { x }_{ n + 1 } - { x }_{ n }) } + { 3 }^{ \frac { 1-2n }{ 2 } }\sin {( { x }_{ n }) } \sin {( x_{ n + 1 }) } = 0

Let x 1 = tan ⁡ − 1 2 3 > x 2 > x 3 ⋯ { x }_{ 1 } = \tan ^{ -1 }{ \dfrac { 2 }{ \sqrt { 3 } } } > { x }_{ 2 } > { x }_{ 3 }\cdots be positive real numbers satisfying the equation above for n ≥ 1 n \ge 1 . Find the value of lim ⁡ n → ∞ x n \displaystyle \lim _{ n\rightarrow \infty }{{ x }_{ n }} .

π 10 \frac {\pi}{10} − π 12 \frac {-\pi}{12} π 6 \frac {\pi}{6} π 8 \frac {\pi}{8}

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Chew-Seong Cheong
Jan 31, 2018

sin ⁡ ( x n + 1 − x n ) + 3 1 − 2 n 2 sin ⁡ x n sin ⁡ x n + 1 = 0 sin ⁡ x n + 1 cos ⁡ x n − cos ⁡ x n + 1 sin ⁡ x n + 3 1 − 2 n 2 sin ⁡ x n sin ⁡ x n + 1 = 0 Divide both sides by sin ⁡ x n sin ⁡ x n + 1 . cot ⁡ x n − cot ⁡ x n + 1 + 3 1 − 2 n 2 = 0 \begin{aligned} \sin (x_{n+1}-x_n) + 3^{\frac {1-2n}2} \sin x_n \sin x_{n+1} & = 0 \\ \sin x_{n+1}\cos x_n - \cos x_{n+1}\sin x_n + 3^{\frac {1-2n}2} \sin x_n \sin x_{n+1} & = 0 & \small \color{#3D99F6} \text{Divide both sides by }\sin x_n \sin x_{n+1}. \\ \cot x_n - \cot x_{n+1} + 3^{\frac {1-2n}2} & = 0 \end{aligned}

⟹ cot ⁡ x n + 1 − cot ⁡ x n = 3 1 − 2 n 2 ∑ k = 1 n cot ⁡ x k + 1 − ∑ k = 1 n cot ⁡ x n = ∑ k = 1 n 3 1 − 2 k 2 cot ⁡ x n + 1 − cot ⁡ x 1 = 3 ∑ k = 1 n 3 − k cot ⁡ x n + 1 − 3 2 = 3 ∑ k = 1 n 3 − k ⟹ cot ⁡ x n + 1 = 3 ∑ k = 1 n 3 − k + 3 2 lim ⁡ n → ∞ cot ⁡ x n + 1 = 3 3 ( 1 1 − 1 3 ) + 3 2 = 3 ⟹ lim ⁡ n → ∞ a n = cot ⁡ − 1 3 = π 6 \begin{aligned} \implies \cot x_{n+1} - \cot x_n & = 3^{\frac {1-2n}2} \\ \sum_{k=1}^n \cot x_{k+1} - \sum_{k=1}^n \cot x_n & = \sum_{k=1}^n 3^{\frac {1-2k}2} \\ \cot x_{n+1} - \cot x_1 & = \sqrt 3 \sum_{k=1}^n 3^{-k} \\ \cot x_{n+1} - \frac {\sqrt 3}2 & = \sqrt 3 \sum_{k=1}^n 3^{-k} \\ \implies \cot x_{n+1} & = \sqrt 3 \sum_{k=1}^n 3^{-k} + \frac {\sqrt 3}2 \\ \lim_{n \to \infty} \cot x_{n+1} & = \frac {\sqrt 3}3 \left(\frac 1{1-\frac 13}\right) + \frac {\sqrt 3}2 = \sqrt 3 \\ \implies \lim_{n \to \infty} a_n & = \cot^{-1} \sqrt 3 = \boxed{\dfrac \pi 6} \end{aligned}

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Very nice. Much easier than finding a formula for tan ⁡ ( x n ) , \tan(x_n), which is what I did!

Patrick Corn - 3 years, 4 months ago

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Glad that you like it.

Chew-Seong Cheong - 3 years, 4 months ago

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Did the same

Aditya Kumar - 3 years, 4 months ago

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