This question comes from the Mathematica documentation. In the documentation of $e$ 1, it shows a neat example of Weyl-type sum:
Letting $$ f(n;r,p) \triangleq \sum_{k=1}^{n} e^{i~r~k^p}, $$ then the sequence ${\left\{f(n;1000e,\frac{1}{2})\right\}}_{n=1}^{2000}$ shows like 2, which contains multiple spirals.
When various $r$ is adopted, the figures become different. Another example is the case for $r=10000\gamma$, where $\gamma$ is the Euler's constant. The result of such a case can be seen in the Neat Example part of 3. Moreover, modification on $p$ has different effects, which can be seen in 4.
However, there are no references in these Mathematica documentations, and I can't find relevant explanation via searching.
A basic explanation of this problem is separating the real and the imaginary parts. Letting $$ g(n;r,p) \triangleq \sum_{k=1}^{n} \cos(r~k^p), h(n;r,p) \triangleq \sum_{k=1}^{n} \sin(r~k^p), $$ we can find multiple plateaus in the figure. For instance, ${\left\{g(n;1000e,\frac{1}{2})\right\}}_{n=1}^{2000}$ has the figure 5. The aligned plateaus of $g$ and $h$ create the spirals in $f$. However, such an explanation does not make much help. I still wonder why these plateaus happen, why the plateaus of $g$ and $h$ are aligned, and what's the relationship between the plateaus and the value of $r$ and $p$.
- This question is partly like that in 6. However, since $k^p$ is not a polynomial, the graph does not have periodicity, and other method should be adopted.