Is $\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\cdots\sin x\cdots\right)\right)=\frac4{\pi}\sum\limits_{k=0}^\infty\frac{\sin(2k+1)x}{2k+1}$?

Solution 1:

I don't think you can easily compute $\mathcal F{\left( f \circ f \right)}$ from $\mathcal F{(f)}$, where $\mathcal F$ is the Fourier transform.

But you can prove the statement above ($f^n$ means $f \circ \cdots \circ f$, $n$-times):

$$ \lim_{n \to +\infty} f^n(x) = \begin{cases} 0 & \text{if } \frac{x}{\pi}\in\mathbb{Z}\\ \mathrm{sign}(\sin{x}) & \text{otherwise} \end{cases}$$

And then deduce:

$$ \lim_{n\to\infty}{f^n(x)}=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin{(2k+1)x}}{2k+1} $$ $f(x) = \sin( \frac{\pi}{2} x) $ sends $[-1,1]$ onto itself, it has $3$ fixed points: $-1, 0$ and $1$.

The derivative at these points is respectively: $0, \frac{\pi}{2} > 1$ and $0$. So $-1$ and $1$ are super-attracting fixed points (any point close to $-1$ or $1$ move closer and closer) whereas $0$ is repelling fixed point (any point close to $0$ move away).

If $0 < x < 1$, $f^n(x) \to 1$ (very fast) and if $-1 < x < 0$, $f^n(x) \to -1$.

I can give more details, if needed.