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Vietnamese Mathematical Olympiad

Algebra Difficulty 7.3 National Olympiad, round 2 Prove it Vietnam

For every couple of real numbers $(a, b)$ , consider the sequence of numbers $\{x_n\}$ , $n = 0, 1, 2, \dots$ , defined by:
$x_0 = a \quad \text{and} \quad x_{n+1} = x_n + b \sin x_n \quad \text{for every } n = 0, 1, 2, \dots$
Prove that:
1) for every real number $a$ , the sequence $\{x_n\}$ corresponding to $(a, b)$ has a finite limit when $n$ tends to infinity. Find this limit.
2) for every number $b > 2$ , there exists a real number $a$ such that the sequence $\{x_n\}$ corresponding to $(a, b)$ does not have a finite limit when $n$ tends to infinity.

Solution

1).
+ For $a = k\pi$ ( $k \in \mathbb{Z}$ ), we have $x_n = k\pi \quad \forall n \in \mathbb{N}$ , therefore $\lim_{n \to \infty} x_n = k\pi$ .

+ For $a \neq k\pi$ ( $k \in \mathbb{Z}$ ), consider the function $f(x) = x + \sin x$ defined on $\mathbb{R}$ . We have $f'(x) = 1 + \cos x \ge 0 \quad \forall x \in \mathbb{R}$ . It follows that:

i) If $a \in (2k\pi; (2k+1)\pi)$ the sequence $\{x_n\}$ is increasing and bounded above by $(2k+1)\pi$ , and it is easy to see that $\lim_{n \to \infty} x_n = (2k+1)\pi$ .

ii) If $a \in ((2k-1)\pi; 2k\pi)$ the sequence $\{x_n\}$ is decreasing and bounded below by $(2k-1)\pi$ , and it is easy to see that $\lim_{n \to \infty} x_n = (2k-1)\pi$ .

Thus, for every $a \in \mathbb{R}$ , the sequence $\{x_n\}$ corresponding to $(a, 1)$ is convergent and $\lim_{n \to \infty} x_n = (2[a/2\pi]+\operatorname{sign}({a/2\pi}))\pi$ .

2).
For $b > 2$ , it is easy to show that there exists $a_0 \in (0; \pi)$ such that $2a_0 = b \cdot \sin a_0$ and the sequence $\{x_n\}$ corresponding to $(a = \pi - a_0, b)$ is a periodic sequence of period 2.

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