# Analytic continuation for PI(1+z^(4^n))

How to do analytic continuation for following function? $$f(z) = \prod_{n=0}^{+\infty} {(1+z^{4^n})}$$ Evidently it satisfies $f(z)f(z^2)=\dfrac{1}{1-z}$...

• Analytic continuation to where? The function seems to have a natural boundary at the unit circle... – Igor Rivin Oct 28 '13 at 15:54
• Oh.. It seems that you're right. – Lwins Oct 28 '13 at 19:30
• Notice that the function has zeros at all the $4^k$th roots of unity -- a dense set. – Igor Rivin Oct 28 '13 at 20:01
• @IgorRivin I don't think that is a very convincing argument -- what about the function $\prod_{n\geq 0} (1+z^{2^n}) = \frac1{1-z}$? – Harry Richman Feb 6 '18 at 3:11

Take a the line $z=t\exp(i\phi)$ with $\phi=2\pi/{4^{n_0}}$ for some $n_0\in \mathbb{N}$. Then $$f(z)=\Pi_{n=0}^{n_0-1}\left(1+t^{4^n}\exp\left(i\phi^{4^n}\right)\right)\Pi_{n=n_0}^{\infty}(1+t^{4^n})$$ Since $4^{n_0}/2\neq 4^{n^\prime}$, $\Pi_{n=0}^{n_0-1}\left(1+\exp\left(i\phi^{4^n}\right)\right)\neq0$, and therefore $f$ diverges at $t=1$.
So your analytic extension would have an accumulation of essential singular points, on the sequence $$\left\{\exp\left(i\frac{2\pi}{4^n}\right),1\right\},$$ and that's not very popular in the analytic community.
It was conjectured by G. Pólya in 1916 and proved by F. Carlson in 1921 that a power series with integer coefficients and radius of convergence one is either rational or has the unit circle as a natural boundary. Since $f(z)$ is clearly not rational, the second alternative must hold.
• See George Pólya, Collected Papers, MIT Press, 1974, pp. 192-198 and 779-780 for information on Carlson's theorem. $f(z)$ is not rational because, for instance, it has arbitrarily long strings of 0 coefficients. – Richard Stanley Oct 28 '13 at 19:52