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The complex power series $\sum_{n=1}^{\infty}\frac{z^{n^2}}{n^2}$ has radius $1$ (Ratio Test) and is absolutely convergent along $|z|=1$. Recalling something that my calculus professor (Ray Mayer, emeritus of Reed College) showed me 15 years ago, I started looking at a "graph" of this function. More precisely, here are plots of the images of $z$ with constant magnitude under this power series.

graph of the fractal

(The mapped curves are the images of $|z|=1$, $|z|=0.9$, and $|z|=0.8$. At the far right you can see $\sum\limits_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6}$.)

So... what the heck is going on with the fractal behavior of the image of the boundary? Is there any understanding of this kind of behavior from power series? For instance, rotating $z$ by some angles might leave you with roughly the original series after a bit of rotation, scaling and translation. But I haven't been able to see how that would all come together.

I have a hunch that the "berries" along the inside of the leaf happen around values of $z$ with interesting arguments, but I haven't sat down to map out what those arguments might be.


EDIT: Indeed, the "leaftips" and "berries" seem to happen at regular $z$ values. Starting at the largest leaftip in quadrant I and moving clockwise, the leaftips are the images of $\exp\left(\frac{\pi}{2}i\right), \exp\left(\frac{\pi}{4}i\right), \exp\left(\frac{\pi}{6}i\right)$,... Similarly, starting from the largest berry and moving clockwise the "seeds" of the berries are the images of $\exp\left(\frac{\pi}{3}i\right), \exp\left(\frac{\pi}{5}i\right), \exp\left(\frac{\pi}{7}i\right)$,...

It appears that the "arc" from the large berry in quadrant IV to the large berry in quadrant I is parametrized by $\exp(it)$ with $t\in\left(-\frac{\pi}{3},\frac{\pi}{3}\right)$. Further, that the large leaf in quadrant I that crosses into quadrant II is parametrized by $t\in\left(\frac{3\pi}{7},\frac{3\pi}{5}\right)$. These two sections (and indeed any of the "leaves", "subleaves", etc.) ought to be similar in the fractal sense.

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    I'm not sure I'm adding anything here, but there is a result - the Hadamard Gap theorem: http://en.wikipedia.org/wiki/Ostrowski%E2%80%93Hadamard_gap_theorem - that if the powers in a power series grow sufficiently quickly, the series is lacunary. One would then expect some sort of highly nondifferentiable behavior on the boundary of the disc of convergence, because otherwise what's the obstruction to analytic continuation? So perhaps it's not so surprising that we get these crazy fractals for a wide class of power series where the powers grow fast?2011-12-22

2 Answers 2

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Let $f(z) = \sum_{n=1}^\infty \frac{z^{n^2}}{n^2}$. Separate this into the parts with odd $n$ and even $n$: $f_o(z) = \sum_{n \text{ odd}} \frac{z^{n^2}}{n^2}$, $f_e(z) = \sum_{n \text{ even}} \frac{z^{n^2}}{n^2} = f(z^4)/4$. Thus $f(z) = \sum_{j=0}^\infty f_o(z^{4^j})/4^j = f_o(z) + \frac{f_o(z^4)}{4} + \frac{f_o(z^{16})}{16} + \ldots$ The plot of $f_o(e^{it})$ for $0 \le t \le 2 \pi$ looks like this:

enter image description here

To produce the plot of $f(e^{it})$, think of a point going around the plot of $f_o$, and then a second describing a similar motion around that point but with 4 times the speed and 1/4 the scale, a third going around the second with $4^2$ times the speed and $1/4^2$ the scale of the original, etc.

The symmetry of the graph of $f_o$ is due to the fact that $f_o(e^{i\pi/4} z) = e^{i\pi/4} f_o(z)$ as well as $f_o(\overline{z}) = \overline{f_o(z)}$.

$f_o$ itself can be separated into parts for $n \equiv 0 \mod 3$ and $n \equiv 1 \text{ or } 2 \mod 3$: if $g = \sum_{n \equiv 1 \text{ or } 5 \mod 6} z^{n^2}/{n^2}$, we have $f_o(z) = \sum_{j=0}^\infty g(z^{9^j})/9^j = g(z) + \frac{g(z^9)}{9} + \frac{g(z^{81})}{81} + \ldots $

The plot of $g(e^{it})$ looks like this:

enter image description here

with symmetry coming from the fact that $g(e^{i\pi/12} z) = e^{i\pi/12} g(z)$.

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    Do you still consider this an answer to the question? I feel that all you have done is passed the buck. You've related the fractal symmetry of $f$ down to the fractal symmetry of $g$, and also explained the $24$-symmetry of $g$. But unless $g$ itself has fractal symmetry, it is not clear to me why the relation between $g$ and $f$ should give $f$ fractal symmetry. After all, the similar relationship between $\ln(z)$ and $\ln(z^k)/k$ does not give $\ln(z)$ fractal symmetry.2012-01-02
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Here's the input circle being slowly warped to the resulting output fractal. gif Code.

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    Sounds neat. I can't see it because I don't have a dropbox account (and I don't want to create one just for this). Could you post it somewhere with more open access? Also, this is probably more appropriate as a comment to the OP. Some users may down vote this as an answer.2015-08-20