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A robot takes steps of length $1$ and turns left $k$ radians before its $k$th step. Will it get arbitrarily close to any chosen point in the plane of it motion?

Suppose it starts at $(0,0)$ facing in the positive $x$ direction. It turns left $1$ radian, then steps forward $1$. Its coordinates, after $n$ steps, can be expressed as:

$$x=\sum_{k=1}^n \cos{\left(\frac{k(k+1)}{2}\right)}$$

$$y=\sum_{k=1}^n \sin{\left(\frac{k(k+1)}{2}\right)}$$

Here is a desmos animation showing its movement. It seems plausible that it will get arbitrarily close to any point, if given enough steps.

My attempt

I considered the facts that

$$\sum\limits_{k=1}^n \cos k = \frac{\cos\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2} \right)}{\sin\left(\frac{1}{2}\right)}$$

$$\sum\limits_{k=1}^n \sin k = \frac{\sin\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2} \right)}{\sin\left(\frac{1}{2}\right)}$$

I tried to derive similar formulas for the robot's coordinates. I got

$$x=\text{Re}(e^{i}+e^{3i}+e^{6i}+...)$$

$$y=\text{Im}(e^{i}+e^{3i}+e^{6i}+...)$$

But this doesn't seem to help.

EDIT: For what it's worth, $\sum\limits_{k=1}^{n} \sin(k^2)$ is unbounded.

EDIT2 Clarification: When I ask whether the robot will get arbitrarily close to any chosen point in the plane of its motion, I am considering only its locations at the end of each step, not the line segments of length $1$ between steps.

Dan
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  • Don't know whether this helps, but substitute $n = \frac{k(k + 1)}{2}$ into the facts given. You might get something. – Yajat Shamji Jan 26 '23 at 13:24
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    It will stay within $50$ for about $100000$ steps then, within $1500$ steps will move about $330$. Then it repeats, but goes off in a different direction. I think that might be related to $103993/33102$, which is a very close approximation for $\pi$ – Empy2 Jan 26 '23 at 14:03
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    @Empy2 At $n=60000$ its coordinates are approximately $(-156.5,336.4)$, so a distance of about $371$ from the origin. – Dan Jan 26 '23 at 14:25
  • Similarly 22/7 being an approximation to $\pi$ seems to be driving some repetition. – Michael Lugo Jan 26 '23 at 15:07
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    You should absolutely look in Hugh Montgomery's "Ten lectures on the interface between analytic number theory and harmonic analysis": you'll find mathematical descriptions of these sums, as well as plots that show the stationary-phase-like phenomenon that @Empy2 mentioned in their comment. – Greg Martin Jan 26 '23 at 23:36
  • Another intuitively plausible outcome is that it is always within some fixed distance from a curve where in polar coordinates $\theta$ is a continuous function of $r$. – aschepler Jan 26 '23 at 23:44
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    Maybe an idea: see how the path resembles a 2D random_walk/brownian_motion/wiener_process, as example the Wiener representation, in 2D is known to almost surely return to the origin, since the origin is an arbitrary point, maybe you could prove that in infinite time in a finite grid every point is going to be visited almost surely, with luck, maybe you could extend it to the reals (but since is not cited in Wikipedia, maybe there is no demonstration). – Joako Jan 31 '23 at 17:14
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    This is sort of a "pseudo-random Brownian motion": The step directions average out to $0$ by Weyl's polynomial equidistribution theorem . For fixed $h \neq 0$, the step directions at time $k$ and $k+h$ also average out to be uncorrelated. However, the step directions at times $k$, $k+1$ and $k+2$ are not at all independent; we always have $\left( e^{i \binom{k}{2}} \right) \left( e^{i \binom{k+1}{2}} \right)^{-2} \left( e^{i \binom{k+2}{2}} \right) = e^{i}$. So we need a proof of recurrence for Brownian motion which only uses pairwise independence of the steps. – David E Speyer Feb 02 '23 at 14:58

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