TITLE: Solve the functional equation $f(4x(1-x))=\sin(\pi f(x))$ to find an invariant measure of a dynamical system $x_{n+1}=\sin(\pi x_{n})$
QUESTION [7 upvotes]: During the investigation of my thesis I found the following problem:
I need find a injective function $f$ such that 
$$f(4x(1-x))=\sin(\pi f(x))\tag{1}$$
and  $f(0)=0$ and $f(1)=1$.
Remark: I have tried to solve this equation but I have only reached a reformulation:
$$
\frac{4(1-2y)f'(4y(1-y))}{\sqrt{1-f^{2}(4y(y-1))}}=\pi f'(y)
.$$
You might think that this problem is not suitable for this site but for me it is important because it arises in trying to address the following situation.
The situation:
 This question is motivated by a problem of Ergodic Theory in my thesis, the problem is find  invariant measure of dynamical system
$$x_{n+1}=\sin(\pi x_{n}).\tag{2}$$
For this purpose it is sufficient to find its invariant density. In this sense, the idea is find a  function $f$ injective such that satisfies (1) with $f(0)=0$ and $f(1)=1$. If that function existed then we have a change of coordinates given by
$$x=f(y).$$
Therefore, substituting in (2) we have $f(y_{n+1})=\sin(\pi f(y_{n}))$, but by (1) we have
$$ f(y_{n+1})=f(4 y_{n}(1-y_{n})) .$$
Since we assume that $f$ injective then
$$y_{n+1}=4y_{n}(1-y_{n}) \tag{3}.$$
But we know the invariant probability density function of dynamical system (3) is the function 
$$\rho(x)=\frac{1}{4\sqrt{x(1-x)}}.$$
Therefore, the invariant probability density function $g$ of dynamical system (1) is the function
$$g(x)=\left|\frac{df(x)}{dx}\right|\rho(x).$$
Therefore, the whole problem is reduced to finding $f$.

REPLY [6 votes]: Following up on Anthony Quas's post, let $g(x) = 4x(1-x)$ and $h(x) = \sin(\pi x)$.  The equations $g^{(n)}(x) = 1/2$ and $h^{(n)}(x) = 1/2$ each have $2^n$ roots in $[0,1]$. If we sort those roots in order as $y^n_1 < y^n_2 < \cdots < y^n_{2^n}$ and $z^n_1 < z^n_2 < \cdots < z^n_{2^n}$, we should have $f(y_i) = z_i$. 
To see that there is unlikely to be a closed form solution, note that $y^n_1$ behaves like $c_1 4^{-n}$ for some constant $c_1$ and $z^n_1$ behaves like $c_2 \pi^{-n}$. So we must have $f(y) \sim c_3 y^{\log(\pi)/\log(4)}$ as $y \to 0$. Similar arguments should show that we have $f(y^n_i+\delta) - z^n_i \sim c \delta^{\log(\pi)/\log(4)}$ as $\delta \to 0$ for every $y^n_i$ (with the constant $c$ depending on $(n,i)$). This is not how a nice function behaves!
Here is a plot, and list of values, for $n=6$:

0.0002     0.0005
0.0014     0.0033
0.0038     0.0079
0.0074     0.0132
0.0121     0.0211
0.0181     0.0282
0.0252     0.0369
0.0335     0.0465
0.0429     0.0604
0.0534     0.0716
0.0650     0.0832
0.0776     0.0942
0.0912     0.1092
0.1058     0.1220
0.1214     0.1375
0.1379     0.1546
0.1552     0.1790
0.1734     0.1974
0.1924     0.2153
0.2121     0.2310
0.2325     0.2505
0.2536     0.2659
0.2752     0.2833
0.2974     0.3012
0.3201     0.3262
0.3432     0.3454
0.3666     0.3651
0.3904     0.3835
0.4145     0.4082
0.4388     0.4292
0.4632     0.4543
0.4877     0.4813
0.5123     0.5187
0.5368     0.5457
0.5612     0.5708
0.5855     0.5918
0.6096     0.6165
0.6334     0.6349
0.6568     0.6546
0.6799     0.6738
0.7026     0.6988
0.7248     0.7167
0.7464     0.7341
0.7675     0.7495
0.7879     0.769
0.8076     0.7847
0.8266     0.8026
0.8448     0.821
0.8621     0.8454
0.8786     0.8625
0.8942     0.878
0.9088     0.8908
0.9224     0.9058
0.9350     0.9168
0.9466     0.9284
0.9571     0.9396
0.9665     0.9535
0.9748     0.9631
0.9819     0.9718
0.9879     0.9789
0.9926     0.9868
0.9962     0.9921
0.9986     0.9967
0.9998     0.9995