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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 $|q|>1$.

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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
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.

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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

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