1

Define $f_1(x) = \sin(x)$. For each $k \geq 1$, define $f_{k+1}(x) = (f_k \circ f_1)(x)$.

Are the properties of the limiting function (if it exists) well-known? That is, as $n$ tends to infinity, how does $f_n$ behave?

Anything you can think of to help me would be greatly appreciated!

User12345
  • 1,331

1 Answers1

2

$f_2(x) =$ sin$($sin$(x))$ on $\mathbb{R}$ is just sin$(x)$ on the open interval $(-1, 1)$. $f_2$ is monotonically increasing, with values $f_2(-1) \approx -0.8414$ and, of course, $f_2(1) \approx 0.8414$.

If we repeat for $f_3$, we get $f_3$ has values between sin $(-0.8414) \gt -0.8414$, and sin$(0.8414) < 0.8414$.

It's clear that whenever $m > n$, we have $Im f_m \subset Im f_n$. But the "peaks" of the function remain at $\{\frac{\pi}{2} + \pi n : n \in \mathbb{Z}\}$.

Also, since $0$ is an attractive fixed point of the function sin$(x)$, we have that $\lim_{n \to \infty} Im f_n = \{0\}$. Thus, your sequence of functions converges to the 0 function everywhere on $\mathbb{R}$.

Visually, at each iteration, the "peaks" get lowered, and they get arbitrarily close to 0 as $n \rightarrow \infty$.

Utku Alhan
  • 166
  • 8
  • Note: Studying the behaviour of an arbitrary real number under repeated iterations of the sine function also gives pointwise convergence to the 0 function, perhaps with less clutter. – Utku Alhan Mar 18 '14 at 15:04
  • I do believe that your comment that 0 is an attractive fixed point assumes the consequent... I had in mind a proof by induction, using something along the lines of what you showed concerning the images. Thanks though! – User12345 Mar 21 '14 at 16:26
  • @Eppo Yes, I just used the fact without proof, but considering $\sin x < x$ for all $x \in (0, \pi$), and the iteration stabilizes at $\sin x = x$, which yields $x = 0$ makes it quite obvious I think. – Utku Alhan Mar 21 '14 at 18:25