Minkowski question-mark function

Example

Every irrational $x \in (0, 1)$ has a unique infinite continued fraction expansion $x = [0; a_1, a_2, a_3, \dots]$ with positive integers $a_k$. The question-mark function $? \colon [0, 1] \to [0, 1]$ is defined at such $x$ by $$?(x) = 2 \sum_{k=1}^{\infty} (-1)^{k+1} \, 2^{-(a_1 + a_2 + \cdots + a_k)}.$$ For a rational $x = [0; a_1, \dots, a_n]$ the same formula with a finite sum is used, and $?(0) = 0$. The value does not depend on which of the two finite expansions of a rational is chosen. For example, $\frac12 = [0; 2] = [0; 1, 1]$ gives $2 \cdot 2^{-2} = \frac12$ and $2(2^{-1} - 2^{-2}) = \frac12$.

Some values: $?(1) = 1$, $?(\frac13) = 2 \cdot 2^{-3} = \frac14$, and for $x = [0; 1, 1, 1, \dots] = (\sqrt5 - 1)/2$ the geometric series gives $?(x) = 2 \cdot \frac{1/2}{1 + 1/2} = \frac23$.

The function $?$ is continuous and strictly increasing, so it is a homeomorphism of $[0, 1]$ onto itself. Its derivative exists and equals $0$ at almost every point of $[0, 1]$. So $?$ is a singular function. Unlike the Cantor function, it is strictly increasing, so it is constant on no interval.

The function maps the rationals of $[0, 1]$ onto the dyadic rationals $m/2^n$ of $[0, 1]$. It maps the quadratic irrationals of $(0, 1)$ onto the rationals that are not dyadic.