I would like to know if there is a closed form for this function:
$ f(x) = \begin{cases} \frac{1}{2}, & \text{if}\ 0 \leq x < \frac{1}{2} \\ \frac{1}{4}, & \text{if}\ \frac{1}{2} \leq x < \frac{1}{2} + \frac{1}{4} \\ \frac{1}{8}, & \text{if}\ \frac{1}{2} + \frac{1}{4} \leq x < \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \\ \frac{1}{16}, & \text{if}\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \leq x < \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \\ ... \end{cases} $
Fig. 1: Graphical representation of function $f$ (formula (0)). The red vertical segments are a plotting artefact.
Here is a solution:
$$y=f(x)=2^{\lfloor\log_2(1-x)\rfloor}\tag{0}$$
where $\lfloor \cdots \rfloor$ is the floor function and $\log_2$ is the logarithm function with base $2$.
Remark: Consideration of the extreme cases of the "integer part of $a$", which can be $a$ or $a-1$, one gets
which are the equations of the lines containing resp. the "upper" and "lower" points of discontinuity of the graphical representation (dotted blue lines in the figures).
Proof of formula (0):
Fig. 2: Graphical representation of function $g$, given by formula (1), related to function $f$ by $f(x)=g(1-x)$.
In fact, it will be equivalent and simpler to work on function $g$ with graphical representation given in Fig. 2 and establish that its equation is
$$y=g(x)=2^{\lfloor\log_2(x)\rfloor}\tag{1}$$
Function $g$ is defined as taking values $$y=\tfrac{1}{2^{n+1}}\tag{2}$$ for
$$\tfrac{1}{2^{n+1}} < x \le \tfrac{1}{2^{n}}\tag{3}$$
Remark: $g$ is in fact a rather well known function giving the highest power of 2 smaller than or equal to a given number $x$.
Let us attempt to express $n$ as a function of $x$.
Applying the increasing function $\log_2$ to this double inequation, we get: $$-(n+1) < \log_2(x) \le -n$$ or, reversing the sign and the direction of inequalities: $$n \le -\log_2(x) < n+1$$ Otherwise said: $$n=\lfloor -\log_2(x) \rfloor.$$ Plugging this expression of $n$ into relationship (2) gives: $$y=2^{-(n+1)}=y=2^{-\lceil -\log_2(x) \rceil}$$ which is equivalent to expression (1) (where $\lceil...\rceil$ is the "ceiling" function).