what shape does this function $\delta_{x}(x)$ look like if it's a fractal what is its fractal dimension?
so $\alpha$ is a real number and $\delta_{\alpha}(x)$ is a function based of $\alpha$ written in base $2$
let's say $\alpha=101.001010100110100010..._2$ then $$\delta_{\alpha}(x)=1+\frac{0x}{1!}+\frac{0x^2}{2!}+\frac{x^3}{3!}+\frac{0x^4}{4!}+\frac{x^5}{5!}+\dots$$
when $\alpha$ in base two has finite digits or an endless string of $1$'s then $\delta_{\alpha}(x)$ isn't defined
my question is what shape does $\delta_{x}(x)$ make when plotted. I know that it's not a continuous function so I wonder if it's a fractal.
We consider the following function:
$$ f(x) = \sum_{n=0}^{\infty} \frac{\lfloor 2^n x \rfloor - 2\lfloor 2^{n-1}x\rfloor}{n!}x^n. $$
This function coincides with $\delta_x(x)$ when $x$ is not a dyadic rational, and so, it suffices to study this function instead. Its graph looks like:
This function turns out to be of bounded variation on any interval, and in particular, the Hausdorff dimension of its graph is exactly $1$. So, although the graph looks fun, I would not consider this as a fractal.
Indeed, if we write
$$ f_n(x) = \frac{\lfloor 2^n x \rfloor - 2\lfloor 2^{n-1}x\rfloor}{n!}x^n, $$
then for each positive integer $m$, the total variation of $f_n$ on $[-m, m]$ satisfies
\begin{align*} V(f_n,[-m,m]) &\leq \frac{1}{n!} \left( \int_{-m}^{m} \left| \mathrm{d}(x^n) \right| + \int_{-m}^{m} m^n \left| \mathrm{d}(\lfloor 2^n x \rfloor - 2\lfloor 2^{n-1}x\rfloor) \right| \right) \\ &\leq \frac{1}{n!}(2m^n + (2m)^{n+1}). \end{align*}
So it follows that
$$ V(f, [-m,m]) \leq \sum_{n=0}^{\infty} \frac{1}{n!}(2m^n + (2m)^{n+1}) < \infty, $$
and the desired claim follows.