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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    $\begingroup$ I should note that this is just one example. This behavior is showing up for similar graphs of $\sum_{n=n_0}^{\infty}\frac{z^{f(n)}}{f(n)}$, where $f$ is other increasing functions from $\mathbb{N}$ to $\mathbb{N}$. So far, this one and the one with $f(n)=\binom{2n}{n}$ are my favorites. $\endgroup$
    – 2'5 9'2
    Nov 23, 2011 at 0:38
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    $\begingroup$ ...and speaking of which, the imaginary part of your sum essentially already is Weierstrass's function. $\endgroup$ Nov 23, 2011 at 3:38
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    $\begingroup$ @J.M. I agree that the known information about Weierstrauss's function will almost certainly help (if not fully) explain this image. In another week I might finally have time to read up on it and I might ask you to link that article as an answer. But I'm also interested in the similar fractal behavior that I see for any $\sum \frac{x^{f(n)}}{f(n)}$ where $f$ grows fast enough for the series to be absolutely convergent when $|x|=1$. $\sum \frac{x^{n!}}{n!}$, $\sum \frac{x^{\binom{2n}{n}}}{\binom{2n}{n}}$, and $\sum \frac{x^{2^n}}{2^n}$ are some examples. $\endgroup$
    – 2'5 9'2
    Dec 5, 2011 at 17:12
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    $\begingroup$ Note that to get the Weierstrass function you actually want to use $f(n)=b^n$ (maybe with a factor of $\pi$). In your base example, with $f(n)=n^2$, you actually get Riemann's "nowhere" differentiable function $$R(\theta)=\sum_{n\ge 0} {\sin\big(n^2\theta\big)\over n^2}$$ This is actually differentiable at points of the form $\pi{2p+1\over 2q+1}$, where $p$ and $q$ are integers. I'm not sure what goes on for other $f(n)$ (e.g. $f(n)=n^{k}$ for $k>2$), though $f(n)=n!$ nearly gives Darboux's function. See this paper for references: epubl.ltu.se/1402-1617/2003/320/LTU-EX-03320-SE.pdf $\endgroup$
    – B R
    Dec 8, 2011 at 19:57
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    $\begingroup$ $f(n)=n$ is actually also... exciting. $\endgroup$ Dec 10, 2011 at 12:18

1 Answer 1


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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    $\begingroup$ Hi Robert - I think this is an excellent breakdown of the major (non-fractal) symmetry down to the function you have dubbed $g$. But have you suggested an explanation for the fractal symmetry? For instance, are you suggesting that this breakdown continues to arbitrarily high degree? And a totally different question: could this "break-down" approach apply to general $\sum\frac{z^{f(n)}}{f(n)}$? $\endgroup$
    – 2'5 9'2
    Dec 22, 2011 at 2:58
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    $\begingroup$ It has to do with the fractal almost-symmetry too. $f(z) = f_o(z) + f(z^4)/4$. So whatever happens at $z= e^{i\theta}$ is echoed, at $1/4$ the scale, at $z = e^{i\theta/4}$, $e^{i(\theta/4+\pi/2)}$, $e^{i(\theta/4 + \pi)}$, and $e^{i(\theta/4 + 3\pi/2)}$. $\endgroup$ Dec 22, 2011 at 4:08
  • $\begingroup$ Couldn't the same thing be said about functions not exhibiting fractal behavior? For instance, $\ln(z)=\ln(z^4)/4$ (for restricted values of $z$) , but along $|z|=1$, I don't believe $\ln$ has any fractal symmetry. So you are relying on the fractal symmetry inherent in your function $f_o$ to "echo" down to smaller scales. $f_o$ in turn seems to get its fractal symmetry from $g$. Can $g$'s fractal symmetry be explained? $\endgroup$
    – 2'5 9'2
    Dec 22, 2011 at 5:42
  • $\begingroup$ 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. $\endgroup$
    – 2'5 9'2
    Jan 2, 2012 at 6:18

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