33. [25] Given a function $f$, let $\pi(f)=f \circ f \circ f \circ f \circ f$. The attached sheet has the graphs of ten smooth functions from the interval $(0,1)$ to itself. The left-hand side consists of five functions:
- $F_{1}(x)=0.005+\frac{1}{2} \sin 2 x+\frac{1}{4} \sin 4 x+\frac{1}{8} \sin 8 x+\frac{1}{16} \sin 16 x+\frac{1}{32} \sin 32 x$;
- $F_{2}(x)=F_{1}\left(F_{1}(x+0.25)\right)$;
- $F_{3}(x)=F_{1}\left((1-x) F_{1}\left((1-x)^{2}\right)\right)$;
- $F_{4}(x)=F_{1}(x)+0.05 \sin (2 \pi x)$;
- $F_{5}(x)=F_{1}(x+1.45)+0.65$.
The right-hand side consists of the five functions $A, B, C, D$, and $E$, which are $\pi\left(F_{1}\right), \ldots, \pi\left(F_{5}\right)$ in some order. Compute which of the functions $\{A, B, C, D, E\}$ correspond to $\pi\left(F_{k}\right)$ for $k=1,2,3,4,5$.
Your answer should be a five-character string containing $A, B, C, D, E$, or $X$ for blank. For instance, if you think $\pi\left(F_{1}\right)=A$ and $\pi\left(F_{5}\right)=E$, then you would answer $A X X X E$. If you attempt to identify $n$ functions and get them all correct, then you will receive $n^{2}$ points. Otherwise, you will receive 0 points.
Answer: $D A C B E$
Solution: First, note that $F_{5}$ and $E$ are the only functions whose range is outside the range $(0,0.7)$, so they must be matched.
For graphs $F_{2}$ and $F_{3}$, split the square $[0,1]^{2}$ into a three by three grid of $\frac{1}{3} \times \frac{1}{3}$ squares. The only squares that the graphs pass through are $\left[\frac{1}{3}, \frac{2}{3}\right) \times I$ and possibly $\left[\frac{2}{3}, 1\right] \times\left[0, \frac{1}{3}\right)$, so after five iterates the range of these graphs stabilizes to $f\left(\left[\frac{1}{3}, \frac{2}{3}\right)\right)$. Graph $F_{3}$ corresponds to the flatter graph $C$, while graph $F_{2}$ to the bumpier graph $A$. This is good enough for 9 points.
It remains to distinguish between $F_{1}$ and $F_{4}$. The details are a bit messy, so here are two ideas:
- Split the grid in a $5 \times 5$ grid of $\frac{1}{5}$-by- $\frac{1}{5}$ squares. It is clear these will distinguish $F_{1}$ and $F_{4}$ since $B$ does not pass through $[0.2,0.4) \times[0.4,0.6)$ but $D$ does not.
- Observing the definitions of $F_{1}$ and $F_{4}$, we have that the phases of the waves that make up $F_{1}$ "agree" much more than they do in $F_{4}$, since $\pi$ is irrational. This suggests that $\pi\left(F_{4}\right)$ will have more chaotic behavior than $\pi\left(F_{1}\right)$.

The final answer is $D A C B E$.
Remark. The graphs were made using gnuplot.