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Calculus Level 3

If a 0 , a 1 , a 2 , … a_0, a_1, a_2, \ldots be a sequence of numbers that satsify the recurrence relation a t + 1 = 1 2 ( 1 + a t ) a_{t+1} = \sqrt{\frac12 (1+a_t)} for t = 0 , 1 , 2 , … t = 0,1,2,\ldots .

Evaluate lim ⁡ n → ∞ cos ⁡ ( 1 − a 0 2 a 1 × a 2 × ⋯ × a n ) \displaystyle \lim_{n\to\infty} \cos \left( \frac{\sqrt{1-a_0^2}} {a_1\times a_2\times \cdots \times a_n} \right ) .

1 a 0 a_0 1/2 0

Ojasee Duble by Ojasee Duble , 19, India

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1 solution

Hosam Hajjir
May 13, 2018

We can assume that ∣ a 0 ∣ ≤ 1 | a_0 | \le 1 , and thus, there exists t 0 t_0 such that a 0 = cos ⁡ t 0 a_0 = \cos t_0 . Using the given recurrence, we have a 1 = 1 2 ( 1 + cos ⁡ t 0 ) = cos ⁡ t 0 2 a_1 = \sqrt{ \frac{1}{2} (1 + \cos t_0 ) } = \cos \dfrac{t_0}{2} . It easy to see that a k = cos ⁡ t 0 2 k a_k = \cos \dfrac{t_0}{2^k} . Therefore, ∏ k = 1 ∞ a k = ∏ k = 1 ∞ cos ⁡ t 0 2 k = sin ⁡ t 0 t 0 \displaystyle \prod_{k = 1 }^{\infty} a_k = \prod_{k=1}^{\infty} \cos \dfrac{t_0}{2^k} = \dfrac{\sin t_0}{t_0} by Viète's infinite product formula. Hence, cos ⁡ 1 − a 0 2 ∏ k = 1 ∞ a k = cos ⁡ sin ⁡ t 0 ( sin ⁡ t 0 t 0 ) = cos ⁡ t 0 = a 0 \cos \dfrac{ \sqrt{1 - {a_0}^2} }{\displaystyle \prod_{k = 1 }^{\infty} a_k} = \cos \dfrac{ \sin t_0 }{ \left( \dfrac{\sin t_0 }{t_0} \right) } = \cos t_0 = a_0

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