# Distribution of zeroes of lacunary functions

In a recent Math Stack Exchange question I asked about the function $$f(z)=\sum_{n=0}^\infty z^{2^n},$$ and was informed of its status is a canonical example of a lacunary series with natural boundary at $|z|=1$. A phenomenon observed by the accepted answer was that this function has a multitude of zeroes within the unit disk; it was speculated but not proven that that this set is in fact infinite.

That raises the following questions, for which I've not been able to find appropriate literature:

• Does $f(x)$ have an infinitude of zeros within the unit circle? How can this be proven?
• How are the zeros distributed? (e.g. how many zeros are found within an annulus $0<a\leq |z|\leq b <1$.)
• How does this generalize to other lacunary functions? I am particularly interested in the case where the base in $f(x)$ is a different positive integer.

The main literature I could find online was a Costin and Huang paper from 2009 entitled " Behavior of Lacunary Series at the Natural Boundary". Unfortunately, I found this paper to be too beyond my level to get much out of it; if the paper is relevant, some exposition on it would be appreciated.

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Not exactly what you're asking, but maybe it's worth pointing out the (obvious) fact that natural boundary unit circle does not imply existence of zeros (consider $e^f$). – Christian Remling Aug 14 '14 at 16:47
@christianremling) True, but that example seems very special: if I were to perturb it by adding a constant $\epsilon$, then there certainly can be zeros (and I would suspect them to be multitudinous.) But in any case, my immediate curiousity is to the function defined above. – Semiclassical Aug 14 '14 at 17:07
(Sorry, only marginally on-topic.) Take $n$ variables $\lbrace x_i\rbrace$ and $n$ variables $\lbrace y_j\rbrace$, and consider the product over all $i,j$ of $f(x_iy_j)$. Then the coefficient of the term for which all variables have degree $2^n-1$ is the number of $n\times n$ Latin squares. I've been carrying this little fact around for decades and would love for someone to tell me that it isn't entirely useless (for asymptotics, for example). – Brendan McKay Aug 15 '14 at 3:01
I have no time to post an answer, but read the fifth from the top paper on math.msu.edu/~fedja/pubpap.html and third from the top paper on math.msu.edu/~fedja/prepr.html. Together they'll give you all you need. – fedja Aug 15 '14 at 14:11

In addition to the unique real zero at $z=-0.658626\ldots$, Mahler in a 1982 paper [On the zeros of a special sequence of polynomials, Math. Comp.] determined, to within eight decimal places, eight complex conjugate pairs of zeros of $f(z)$. This gives a total of $17$ fairly precisely located zeros. At the end of that paper he conjectures that there are infinitely many; in fact, in an earlier paper quoted [5] in loc. cit. (On a special function, J. Number Theory), he says that he expects every point on the unit circle to be a limit point of zeros of $f$. I do not know if his conjecture has since been proved or disproved.
In a first approximation, one could look at the zeros of the polynomial truncations of $f$. Then it may help to know that those are equidistributed near the unit circle, a fact not noted by Mahler in his paper. (On page 211, he simply writes: "It seems that the arguments of the zeros are much more uniformly distributed over the values from $0$ to $360$ degrees." He notes instead the weaker statement that, by a general theorem of Jentzsch for power series having a natural boundary, every point of the boundary circle is a limit point of zeros of the truncations.) This follows by a theorem of Erdos and Turan; the same is true for any sequence of polynomials of degree $d \to \infty$ whose leading and free coefficients have non-zero absolute values at least $1$ and whose lengths (sums of the absolute values of all coefficients) are at most $e^{o(d)}$. For integer polynomials, this theorem has been refined by Yuri Bilu. For an exposition of those results and a sketch of their proofs I can refer you to Granville's article The distribution of roots of a polynomial in the volume Equidistribution in Number Theory, An Introduction (Edited by A. Granville and Z. Rudnick).
Mahler was especially fond of this function; for instance, he was able to prove that it takes transcendental values at all algebraic arguments other than $z = 0$. For a sketch of the proof of this, focused on the example of the irrationality of $f(2/3)$, I can refer you if you would be interested to Masser's article Heights, transcendence, and linear independence on commutative group varieties in LNM 1819, which on page 171 notes also a rather curious "functional near-equation" for real arguments $0 < z< 1$. By the way, all known properties of $f$ extend also to the series $\sum z^{b^n}$. – Vesselin Dimitrov Aug 14 '14 at 18:35
You might find additional references to $\sum_n z^{2^n}$ under the name "Fredholm series", which is ironic since Fredholm never studied it. (Fredholm studied the superficially similar series $\sum_n z^{n^2}$, but the name got attached and once attached, it is hard to remove.) – Jeffrey Shallit Aug 14 '14 at 22:04
@VesselinDimitrov: Interesting! Transcendence theory isn't really in my wheelhouse, but I'll take a look. Though I do see one obvious corollary: Zero is algebraic, so the zero-set of $f(z)$ has to be transcendental. – Semiclassical Aug 15 '14 at 2:03