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We can see intuitively that $ f(x)=\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\cdots\sin{x}\cdots\right)\right)\right) $ is the square wave with period $2\pi$ and has the value $0$ at the jumps, i.e $ f(x)= \begin{cases} 0 & \text{if } \frac{x}{\pi}\in\mathbb{Z}\\ \mathrm{sign}(\sin{x}) & \text{otherwise} \end{cases} $ Look at this graph of $x$ and $\sin{\frac{\pi}{2}x}$ to see why :

Graph of x and sin(pi/2*x)

But $f(x)$ is then also exactly equal to the Fourier series of the square wave with period $2\pi$ since Dirichlet conditions assure that the series converges to $0$ (the midpoint) at the jumps as do $f(x)$.

Hence we might be able to show that, $ \sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\cdots\sin{x}\cdots\right)\right)\right)=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin{(2k+1)x}}{2k+1} $

Does anyone have an idea of how to prove this directly? Are there other Fourier series that are equal to a recursive formula of trigonometric functions?

Restated, the problem is to show that if $ f_0(x)=\sin{x},\quad\text{and}\quad f_n(x)=\sin{\left(\frac{\pi}{2}f_{n-1}(x)\right)} $ then, $ \lim_{n\to\infty}{f_n(x)}=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin{(2k+1)x}}{2k+1} $

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    @joriki: late but... J.M.'s (well Töpfer's) paper at [gdz](http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN235181684_0117&DMDID=DMDLOG_0009&IDDOC=37617)2012-08-28

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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.