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.

  • $\begingroup$ 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 $e^f$). $\endgroup$ Aug 14, 2014 at 16:47
  • $\begingroup$ @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. $\endgroup$ Aug 14, 2014 at 17:07
  • $\begingroup$ (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). $\endgroup$ Aug 15, 2014 at 3:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – fedja
    Aug 15, 2014 at 14:11

1 Answer 1


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).

  • $\begingroup$ Given how old this series is (the paper by Costin linked in my question indicates it was studied by Jacobi) I'm rather surprised that the question of infinitude of zeros hasn't been resolved. I'd be curious whether there's been any major advances since Mahler 1982. $\endgroup$ Aug 14, 2014 at 18:22
  • $\begingroup$ 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 $\sum z^{b^n}$. $\endgroup$ Aug 14, 2014 at 18:35
  • $\begingroup$ 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.) $\endgroup$ Aug 14, 2014 at 22:04
  • $\begingroup$ @JeffreyShallit) That's a pretty funny instance of Arnold's principle right there. Following that lead brings up this Flajolet 1987 paper. While it doesn't talk about Fredholm series in detail, it does mention in passing that Mellin transform techniques can be applied to obtain the asymptotic expansion. $\endgroup$ Aug 15, 2014 at 1:55
  • $\begingroup$ @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. $\endgroup$ Aug 15, 2014 at 2:03

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