# What is the Riemann surface of $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$?

The following appears as the second-to-last problem of Stewart's Complex Analysis:

Describe the Riemann surface of the function $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$.

This problem intimidated me when I first saw it as an undergrad, as the series under the root isn't the expansion of any function I know (then or now). That renders the Riemann surface more subtle than usual, since any info about poles and zeroes must be found from this series alone. For instance, it has a zero at $z=0$ and therefore a branch point. But there should be other zeroes — indeed, infinitely many zeroes— since the number associated with any truncation of the series grows as $2^N$!. This suggests that the Riemann surface of this function must be exotic.

I'll give a sketch of an argument below, taking the pedestrian approach of hunting poles and zeroes of the function. However, I'd be glad to see an elegant and advanced perspective of the problem, particularly if it uses some ideas which I wouldn't have encountered as an undergraduate. The more food for thought, the better!

EDIT: As noted by O.L.'s answer below, the series above is the canonical example of a lacunary function with the unit circle as natural boundary. What that leaves open is the nature of the resulting Riemann surface, which would seem to hinge upon the quantity of the function's zeros within the unit circle. David Speyer has given an argument in his answer that this number is at least 10. Is there a more definitive result on the zeroes of this lacuuary function?

• I get $72$ roots in the disk $|z| < 0.999$ and $562$ in the disk $|z| < 0.9999$. I think there are infinitely many roots (in any neighbourhood of a point on the unit circle there should be a root) and I would love to have an asymptotic behaviour of the number of roots in the disk of radius $1-\varepsilon$ – mercio Aug 11 '14 at 11:52
• @mercio As would I. I suspect that the infinitude of the zeros, at least, can be ascribed to the natural boundary at $|z|=1$ acting like a line of essential singularities (which by Picard's theorem display almost all complex values infinitely often in their nbhd). But I don't know what the correct statement is. – Semiclassical Aug 11 '14 at 12:49

## 3 Answers

Truncating series is a good way to hunt for roots, you just have to do it carefully, as Micah points out. I'll show that there are at least $10$ roots in the unit disc. I would guess there are infinitely many.

Set $$f(z) = z+z^2+z^4+z^8+\cdots \quad \mbox{and} \quad f_0=z+z^2+z^4+\cdots+z^{256}$$ According to Mathematica, $|f_0(z)| > 0.066$ on $|z|=0.99.$

f0 = z + z^2 + z^4 + z^8 + z^16 + z^32 + z^64 + z^128 + z^256;
Minimize[{Abs[f0 /. z -> 0.99 E^(I*t)],  0 <= t <= 2 Pi}, t]


On the other hand, $0.99^{512} < 0.00583$ and $$|f(z)- f_0(z)| \leq |z|^{512} + |z^{1024}| + |z^{2048}| + \cdots < \sum_{k=1}^{\infty} |z|^{512 k} < 0.00583/(1-0.00583) < 0.00586.$$ So, by Rouche's theorem, $f(z)$ and $f_0(z)$ have the same number of roots inside $|z|=0.99$.

Again, according to Mathematica, $f_0(z)$ has $10$ roots inside this circle:

SortBy[Select[z /. NSolve[f0 == 0, z], Abs[#] < 0.99 &], Abs]


Namely

{0., -0.658627, 0.120315 - 0.934606 I, 0.120315 + 0.934606 I, -0.685206 - 0.670534 I,   -0.685206 + 0.670534 I, 0.184591 - 0.958351 I, 0.184591 + 0.958351 I, 0.391863 - 0.898257 I, 0.391863 + 0.898257 I}


(The SortBy command tells Mathematica how to sort the output; I ordered in increasing absolute value.) I'm not sure how you'd do this without technology, Stewart may have expected something less precise?

• I imagine you're right about Stewart's intentions being rather more limited. I'll admit, I'm surprised that the roots in the roots persist in the $N\to\infty$ limit. – Semiclassical Aug 8 '14 at 1:00

This answer is closer to a sketch than a proof. If you notice gaps/mistakes/nonsense in this presentation, please let me know...

We want to examine the behavior of the function $f(z)$ inside the square root. It evidently vanishes linearly at the origin. Indeed, for all $|z|<1$ this series is convergent by comparison with the geometric series. If $|z|>1$, however, the series diverges since the terms do not tend to zero. Hence the unit disk is the disk of convergence.

Compared to this straightforward picture, the behavior for $|z|=1$ is more exotic. Consider first a rational rotation, i.e. $z=e^{i\phi}$ for some $\phi\in 2\pi\mathbb{Q}$. Then $2^n\phi$ is either $0$ or $\phi$ modulo $2\pi$ for some sufficiently large integer $N$. (This is equivalent to the binary representation of any rational number eventually terminating or becoming periodic.) In the former case, it follows that $z^{2^n}=1$ for $n\geq N$. For the latter, $z^{2^n}$ is a certain periodic sequence of roots of unity; since this sequence does not include 1, the sum of each period is nonzero and identical. In both cases we may conclude that the series diverges. So every rational rotation gives a singularity.

On the other hand, if $\phi/2\pi$ is irrational, then the sequence $2^n\phi$ modulo $2\pi$ is neither periodic nor terminating: therefore the orbit of $2^n\phi$ will be dense in $[0,2\pi]$. As a consequence, the series sums to zero. So every irrational rotation corresponds to a zero.

Therefore it would appear that the unit circle is interwoven with (countably many) singularities and (uncountably many) zeroes. What does this entail for $y=\sqrt{f(z)}?$ Here my understanding goes from loose to shaky. In the standard case, each zero of $f(z)$ would be a branch point of the corresponding Riemann surface. Consequently one would seem to have a Riemann surface of (uncountably) infinite genus.

Under the (generous) assumption that this is all sound, there are questions which linger for me:

• What is the nature of the singularities at commensurate phases? (Are they poles of some multiplicity, or are they essential singularities?)
• While there are uncountably many zeroes, there is a symmetry under complex conjugation for all roots save $z=0$. Consequently the 'number' of such zeroes is even and we expect to be able to draw branch cuts between all of those on the unit circle. This leaves the branch point at zero; usually I would conclude that there will be a branch point at infinity, but I'm not sure that holds here.
• If one truncates the series expansion at some $2^n$, then the number of zeroes is finite. Most of them are in the neighborhood of the unit circle as would be expected. However, there is also a zero for some $z\in(0,1)$ (as follows from an intermediate value argument.) What happens to this negative real zero when we go to the infinite series?
• The series cannot converge for any $z$ with $|z|=1$, as its terms don't go to zero. If $z$ represents some irrational rotation, that simply means that it diverges by oscillation instead of diverging to infinity... – Micah Aug 7 '14 at 2:47
• @Micah:; Drat, that's an obvious point. So the whole rational v. irrational comparison is simply unnecessary here. Though it's strange: any truncation of the series has $2^{N-1}$ zeroes other than $z=0$. So I don't quite grok what's happening at the unit circle. – Semiclassical Aug 7 '14 at 2:51
• Zeroes of the partial sum need not pass to the limit (e.g., the Taylor polynomials for $e^z$ have lots of zeroes!). I don't have anything like a proof of this, but I have a sneaking suspicion that there are only two branch points ($z=0$ and the negative real one). – Micah Aug 7 '14 at 3:13
• @Micah: Another quite valid point. Without having a functional form, I suppose I was reaching for something to hold on to. And looking at it more carefully, the negative real zero remains on $(-1,0)$ (and seems to converge to a specific spot) as $N\to\infty$. And the intermediate value argument on $(0,1)$ holds for the full series, so that seems fine. – Semiclassical Aug 7 '14 at 3:42
• The series $f(z)=\sum_{k=0}^{\infty}z^{2^k}$ is a canonical example of a function having natural boundary (the so-called lacunary series). – Start wearing purple Aug 8 '14 at 13:49

This is an extended comment to the answer of Semiclassical.

The series $$f(z)=\sum_{k=0}^{\infty}z^{2^k}$$ is a canonical example of a function having natural boundary (the so-called lacunary series). Here the natural boundary is given by the unit circle $|z|=1$, inside which the series is absolutely convergent.

A more general statement is the Ostrowski–Hadamard gap theorem:

Let $\{p_k\}$ be a sequence of positive integers satisfying $\lambda p_k<p_{k+1}$ for any $k\in\mathbb{N}$ with some $\lambda>1$, and let $\{a_k\}$ be a sequence of complex numbers such that the radius of convergence of the series $f(z)=\displaystyle\sum_{k\in\mathbb{N}}a_kz^{p_k}$ is equal to $1$. Then no point on the unit circle $|z|=1$ is a regular point of $f(z)$.

The answers to these MO questions (1, 2) may also be of interest.

• Thanks, that helps. What remains unclear to me is the distribution of zeros of the lacunary series, and from this the resulting Riemann surface. – Semiclassical Aug 8 '14 at 14:35
• @Semiclassical Yes, exactly. Actually I don't understand what kind of answer Stewart may have expected. Experimentally it seems that there is an infinite number of zeros inside the unit circle. Assuming them to be simple, which sounds reasonable, we will obtain something like infinite-genus hyperelliptic curve. – Start wearing purple Aug 8 '14 at 15:03
• That makes sense, with the provision that the natural boundary means one has a double cover of the open unit disk rather than the Riemann sphere. (Though I'd prefer to have a proof/reference regarding the infinitude of zeros.) – Semiclassical Aug 8 '14 at 21:10