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

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    $\begingroup$ Analytic continuation to where? The function seems to have a natural boundary at the unit circle... $\endgroup$
    – Igor Rivin
    Commented Oct 28, 2013 at 15:54
  • $\begingroup$ Oh.. It seems that you're right. $\endgroup$
    – Lwins
    Commented Oct 28, 2013 at 19:30
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    $\begingroup$ Notice that the function has zeros at all the $4^k$th roots of unity -- a dense set. $\endgroup$
    – Igor Rivin
    Commented Oct 28, 2013 at 20:01
  • $\begingroup$ @IgorRivin I don't think that is a very convincing argument -- what about the function $\prod_{n\geq 0} (1+z^{2^n}) = \frac1{1-z}$? $\endgroup$
    – Harry Richman
    Commented Feb 6, 2018 at 3:11

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All the answers above are absolutely fine. Just to give a completely obvious argument:

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.

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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.

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  • $\begingroup$ Could you please give more details about this? I'm just puzzled.. $\endgroup$
    – Lwins
    Commented Oct 28, 2013 at 18:23
  • $\begingroup$ 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. $\endgroup$
    – Richard Stanley
    Commented Oct 28, 2013 at 19:52

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