amparvardi wrote: For all rational $x$ satisfying $0 \leq x < 1$, f is defined by \[f(x)=\{\begin{array}{cc}\frac{f(2x)}{4},&\mbox{ for }0 \leq x < \frac 12,\\ \frac 34+ \frac{f(2x - 1)}{4}, & \mbox{for } \frac 12 \leq x < 1.\end{array}\] Given that $x = 0.b_1b_2b_3 \cdots $ is the binary representation of $x$, find $f(x).$ $f(0.b_1b_2b_3 \cdots )=0.b_1b_1b_2b_2b_3b_3 \cdots$

By above definition of $f$, we have $f(0.b_1b_2...)=\frac{3b_1}{4} + \frac{f(0.b_2b_3...)}{4}=0.b_1b_1 + \frac{f(0.b_2b_3...)}{4}$. So the pattern is like this $f(0.b_1b_2b_3...)=0.b_1b_1b_2b_2...b_nb_n + \frac{f(0.b_{n+1}b_{n+2}...)}{2^{2n}}$. We know that the binary representation of every rational number is periodic. So for rational numbers we get $f(x)=f(0.\overline{b_1b_2...b_n})$. So from the above recursion, $f(x)=0.b_1b_1...b_nb_n + \frac{f(x)}{2^{2n}}$. This means, $f(x)=\frac{2^n(0.b_1b_1.....b_nb_n}{2^n-1}=0.\overline{b_1b_1.....b_nb_n}$. Partially done.