Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME Prove it Ukraine

The sequence $(u_n)$ is defined by conditions: $u_0 = 0$ , $u_1 = \frac{1}{3}$ and $\frac{1}{2} u_n = \frac{1}{2}(u_{n+1} + u_{n-1})$ . Prove that for all positive integers $n$ it holds $|u_n| \le 1$ .

Solution

Let an angle $\varphi \in (0, \frac{\pi}{2})$ such that $\sin \varphi = \frac{\sqrt{5}}{3}$ . Then $\cos \varphi = \frac{2}{3}$ and from the recurrent relation we have: $u_{n+1} = 2u_n \cos \varphi - u_{n-1}$ . Moreover, $u_1 = \sin \varphi$ , $u_2 = \frac{2}{\sqrt{5}} \sin \varphi \cos \varphi = \frac{1}{\sqrt{5}} \sin 2\varphi$ .

Suppose that for every $n \in \mathbb{N}$ it holds: $u_n = \frac{1}{\sqrt{5}} \sin n\varphi$ . Then by mathematical induction we can get:

$u_{n+1} = \frac{2}{\sqrt{5}} \sin n\varphi \cos \varphi - \frac{1}{\sqrt{5}} \sin(n-1)\varphi = \frac{1}{\sqrt{5}} \sin(n+1)\varphi$ , which is required.

The proof follows from the properties of sine function.

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