Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is claimed on it's Wikipedia page that Euler's function, defined by the infinite product $\prod_1^\infty(1-q^n)$ for $|q|<1$, cannot be analytically continued outside the unit disc, that is, the unit disc is a natural boundary of the function. Unfortunately no proof is referred to.

I would like to know how this claim is proved, as it appears to me that it is defined uniquely for $|q|>1$.

share|improve this question
If $a$ is a root of unity, $a^n=1$, then the radial limit of $\prod_1^\infty(1-q^n)$ as $q \to a$ should be zero. So any alleged analytic extension must vanish on all roots of unity, a set dense in the circle. –  Gerald Edgar Nov 1 '12 at 14:42
Indeed. But does that preclude the existence of a unique extension outside the disc? –  Kevin Smith Nov 1 '12 at 15:42
It precludes an extension analytic at any point of the unit circle. In that case, in what sense is a function outside the circle "an extension" of one inside the circle. (It is true that the literature on transseries and analyzable functions may include some cryptic remarks about extending a function beyond a natural boundary. But I haven't figured out how that is supposed to work.) –  Gerald Edgar Nov 1 '12 at 17:59
Thanks Gerald, and Harm. Your responses are very helpful. The sense in which I mean analytic continuation is the existence of a second definition of the function that is convergent both in- and outside the circle, but not necessarily on it. This is a slightly weaker notion, that becomes the ordinary notion of analytic continuation when it is mereomorphic on the boundary. In this case, the series $$\sum \frac{n^2q^n}{1-q^n}^2$$ converges everywhere except on the circle, and agrees with $$\left(q\frac{d}{dq}\right)^2\log f(q)$$ on the open disc, where $f$ is Euler's function. –  Kevin Smith Nov 1 '12 at 22:40
The square in the first expression is supposed to be of the denominator! –  Kevin Smith Nov 1 '12 at 22:42

1 Answer 1

up vote 5 down vote accepted

If an analytic function $f(q)$ is defined on some open neighborhood of $q=a$, and there exists a sequence $a_1,a_2,\dots$ in this neigborhood with $\lim_{n\to\infty}a_n=a$ and $f(a_n)=0$ for all $n$, then the function $f$ must be identical to $0$. Since the zeroes lie dense on the unit circle, $f$ cannot be defined on any open neigborhood of any point on the unit circle. So there is no analytic continuation of the function that can ``escape'' from the unit disc.

share|improve this answer
Please see my comments above, Harm. In the classical sense you have answered my question. I shall reformulate it with a better definition. –  Kevin Smith Nov 1 '12 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.