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I'm not able to mathematically prove that the equation
$$x(k+1)=\sin(x(k))$$
converges asymptotically to zero.
By a simple thought it can be concluded that for any $x(0) \in \mathbb{R}$ it applies
$x(1)=\sin(x(0)) \qquad \qquad \quad x(1) \in [-1,1]$
$x(2)=\sin(\sin(x(0))) \qquad \, \, \, \, \, x(2) \in [-0.84,0.84]$
$x(3)=\sin(\sin(\sin(x(0)))) \, \, \, \, x(3) \in [-0.74,0.74]$
...
Which means that with every step the solution is closer and closer to zero, however I'm not able to prove it.
Does anyone know how to solve this problem please?

Hanul Jeon
  • 27,376

1 Answers1

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This is a great exercise in learning how to apply the tools you learn in your calculus course.

I'm going to limit myself to $x(1) \geq 0$: the case $x(1) \leq 0$ is similar.

Your first observation: that $x(n)$ is getting closer to zero. You have an intuitive idea that $\sin t \leq t$ whenever $t \geq 0$, correct?

And why do you feel this? If your answer sounds like "$\sin t$ simply doesn't grow as fast as $t$", that's a huge clue: use derivatives to prove it.

Now, you get to use another major tool: you have a sequence that is monotone and bounded. Therefore, it has a limit.

Finally, now that you know the limit exists you get to use the calculus of limits to obtain

$$ L = \lim_{n \to +\infty} x(n) $$

How to obtain this limit? Well, the "only" thing you know about $x(n)$ is that $x(n) = \sin(x(n-1))$: what happens if you plug that in?