Find the recurrence.

Geometry Level 4

Let a a and b b ( with a < b a < b ) be positive numbers satisfying

a 1 = a + b 2 , b 1 = a 1 b a_1 = \dfrac{a+b}{2}, \ b_1 = \sqrt{a_1 b}

Let a n = a n − 1 + b n − 1 2 a_n = \dfrac{a_{n-1} + b_{n-1}}{2} and b n = a n b n − 1 b_n = \sqrt{a_n b_{n-1}} for n ≥ 2 n \ge 2 .

Show that lim ⁡ n → ∞ a n = lim ⁡ n → ∞ b n \displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n . What is the value of this limit?

b 2 − a 2 arccos ⁡ ( b a ) \dfrac{\sqrt{b^2-a^2}}{\arccos \left(\frac{b}{a}\right)} b 2 − a 2 arccos ⁡ ( a b ) \dfrac{\sqrt{b^2-a^2}}{\arccos \left(\frac{a}{b}\right)} a 2 − b 2 arccos ⁡ ( a b ) \dfrac{\sqrt{a^2-b^2}}{\arccos \left(\frac{a}{b}\right)} a 2 − b 2 arccos ⁡ ( b a ) \dfrac{\sqrt{a^2-b^2}}{\arccos \left(\frac{b}{a}\right)}

Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Mark Hennings
Sep 19, 2017

If we write a = b cos ⁡ θ a = b\cos\theta , then it is a simple induction to show that a n = b n cos ⁡ ( θ 2 n ) b n = b ∏ j = 1 n cos ⁡ ( θ 2 j ) n ∈ N a_n \; = \; b_n \cos\big(\tfrac{\theta}{2^n}\big) \hspace{1cm} b_n \; = \; b\prod_{j=1}^n \cos\big(\tfrac{\theta}{2^j}\big) \hspace{2cm} n \in \mathbb{N} Moreover it follows that 2 n sin ⁡ ( θ 2 n ) b n = b 2 n sin ⁡ ( θ 2 n ) × ∏ j = 1 n cos ⁡ ( θ 2 j ) = b sin ⁡ θ 2^n \sin\big(\tfrac{\theta}{2^n}\big)\,b_n \; = \; b2^n \sin\big(\tfrac{\theta}{2^n}\big) \times \prod_{j=1}^n \cos\big(\tfrac{\theta}{2^j}\big) \; = \; b\sin\theta and hence that b n = b sin ⁡ θ 2 n sin ⁡ ( θ 2 n ) → b sin ⁡ θ θ = b 2 − a 2 cos ⁡ − 1 a b n → ∞ b_n \; = \; \frac{b\sin\theta}{2^n \sin\big(\tfrac{\theta}{2^n}\big)} \; \to \; \frac{b\sin\theta}{\theta} \; = \; \frac{\sqrt{b^2-a^2}}{\cos^{-1}\tfrac{a}{b}} \hspace{2cm} n \to \infty It is clear that a n a_n also converges to the same limit.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Since b>a only 2nd or forth can be the solution. Cos is never greater than 1. So the solution is 2nd only.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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