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A few months ago, while attempting to create a parameterization of the Hilbert curve, I discovered an interesting function, given by the summation...

$$f(x)=\sum_{n=1}^\infty \frac{\text{sgn}\left(\cos(nx)\right)}{n}$$

...where $\text{sgn}$ is the signum function...

$$\text{sgn}(x)=\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}$$

The function is noteworthy for being nowhere differentiable and nowhere continuous. I have found it nigh impossible to evaluate the function for any $x$ save integer multiples of $2\pi$ (where the sum diverges).

The graph of the function (shown below) seems to be a self-similar fractal, but I don't know enough about fractals to verify this. Even if I did, I'm not sure what I would do with this information.

enter image description here

Anyway, how can I find the value of $f(x)$ for arbitrary $x$?

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    $\begingroup$ I don't know how to answer this question, but I must say this plot looks amazing. $\endgroup$ – Wojowu May 22 at 18:04
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For rational multiples of $\pi$, $\text{sgn}(\cos(nx))$ is periodic and the sum can be done in "closed form" in terms of the Digamma function. Thus for $x = 3\pi/4$, Maple tells me:

sum(sum(signum(cos(k*3*Pi/4))/(k+8*m),k=1..8),m=0..infinity);

$$ \frac{\pi}{4} - \frac{\ln(2)}{4} + \frac{\Psi(1/8) - \Psi(3/8)}{4} $$

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  • $\begingroup$ This is precisely what I thought when I read the post. The problem seemed very straightforward for rational multiples of $\pi$. It seems like other forms of $x$ would be much harder, if not completely intractable. $+1$ $\endgroup$ – Clayton May 22 at 17:18
  • $\begingroup$ Thanks! Do you think I should I try approximating $f(x)$ for other $x$ by taking the limit of a sequence of rational multiples of $\pi$ which converges to $x$? $\endgroup$ – R. Burton May 22 at 17:20
  • $\begingroup$ In principle yes, but the discontinuities may make it hard to control the approximations. $\endgroup$ – Robert Israel May 22 at 17:30
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This is more like an extended comment. From the Fourier series

$$ \operatorname{sign}(\cos(x)) = \frac{\pi}{4}\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} \cos((2k+1)x), $$

we heuristically compute that

\begin{align*} f(x) &= \frac{4}{\pi}\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac{(-1)^k}{n(2k+1)}\cos(n(2k+1)x) \\ &= \frac{1}{\pi} \sum_{l=1}^{\infty} \frac{r_2(l)}{l} \cos(lx) \\ &= \frac{1}{\pi} \sum_{\mathrm{k} \in \mathbb{Z}^2 \setminus\{0\}} \frac{\cos(|\mathrm{k}|x)}{|\mathrm{k}|}, \end{align*}

where $r_2(\cdot)$ is the sum of squares functions and the second step follows by rearranging the sum according to the value of $l = n(2k+1)$, although establishing the validity of such manipulation would be a non-trivial task.

The following is the plot of of the partial sum $\frac{1}{\pi} \sum_{l=1}^{1000} \frac{r_2(l)}{l} \cos(lx)$ evaluated at points of the form $x = \frac{k \pi}{1025}$ for $|k| \leq 1025$, replicating OP's figure.

enter image description here

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