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I was considering the function $f(z) = z+ z^2 + z^4 + z^8 + \ ... \ = \sum_{n=0}^{\infty} z^{2^n} $

It's well known this has a natural boundary at $|z| = 1$ and cannot be analytically continued beyond this boundary.

It naturally follows then that the image of this function over the domain of the open unit disc is going to be all of $\mathbb{C}$.

Now the question: is this mapping a bijection?

I was inspired to ask this since I cannot seem to find a non trivial zero of this function other than $z=0$.

One idea is to try to generalize the ideas in this question's answer from the world of differential equations to generic functional equations (note that $f$ can be characterized by $f(z^2) + z = f(z)$. But I wasn't sure what is the best way to generalize the wronkskian testing procedure to arbitrary functional equations.

Sidharth Ghoshal
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    It is not a bijection, by Liouville's theorem. – Angina Seng Dec 02 '19 at 16:40
  • on general principles like Riemann mapping theorem, it obviously cannot be a bijection – Conrad Dec 02 '19 at 16:41
  • how could one find a concrete counter example? – Sidharth Ghoshal Dec 02 '19 at 16:42
  • Note that $f'(0)=1$. Thus, by Bloch's theorem, the function is invertible at least on $|z|<\frac{\sqrt{3}}{4}+2\times 10^{-4}$. –  Dec 02 '19 at 16:46
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    Why would it "naturally follow" that the image of this function is all of $\mathbb{C}$? I'm pretty sure that there are examples of functions which are continuous on the closed unit disk, holomorphic in the interior but which do not admit an analytic continuation beyond the unit disk. – Bruno Krams Dec 02 '19 at 16:57
  • @BrunoKrams I think I read a theorem once a while back, that the image of a function within its natural boundary is always all of C with up to a finite number of missing points. So I was just going by that intuitively – Sidharth Ghoshal Dec 02 '19 at 19:22
  • @frogeyedpeas I've looked it up in the meantime. Seems to me that the Ostrowski-Hadamard Theorem (https://en.wikipedia.org/wiki/Ostrowski%E2%80%93Hadamard_gap_theorem) yields a contradiction. Just pick a sequence of integers as demanded by the theorem and choose the $(a_j)_{j \in \mathbb{N}}$ such that the resulting gap power series converges for $ \vert z \vert = 1$. The resulting function should satisfy the requirements from my first comment. – Bruno Krams Dec 02 '19 at 19:52
  • @frogeyedpeas It's not hard to construct a Blaschke product $B$ in the disc whose zeros accumulate at every point on the unit circle. Such a function has the unit circle as its natural boundary, and $B(\mathbb D)\subset \mathbb D.$ – zhw. Dec 02 '19 at 21:14
  • you can always find an analytic function on the disc such that it and all its derivatives converge to an infinitely differentiable function and its derivatives on the unit circle and still that is the natural boundary, so you can have natural boundary the unit circle, but the holomoprhic function extending not only continuosly but infinitely differentiable (on the circle of course, so in a "real" one dimensional sense) here – Conrad Dec 02 '19 at 23:32
  • One way to find a non-trivial zero of $f$ can be the following. Note that the expansion of $1/f(z)$ in (positive) powers of $z$, except for the term $1/z$ for the pole at $z=0$, converges on a punctured disc ${0<|z|<r}$ for an $0<r<1$. On the circle $|z|=r$ we have a pole of $1/f$ and therefore a non-trivial zero of $f$. With this series we are fortunate that the coefficients of the series of $1/f$ alternate. This implies that there is a singular point at $z=-r$. So $z=-r$ is a pole of $1/f$ and a zero of $f$. – conditionalMethod Dec 03 '19 at 04:15
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    I haven't seen if there is a nice expression for $-r$, but you can express it as a limit $1/\limsup_n(b_n)^{1/n}$ where $b_n$ satisfies $b_0=1$ and $b_n=-\sum_{k=0}^{n-1}b_na_{n-k}$, where $a_n=1$ for $n$ a power of $2$ minus $1$ and zero otherwise. – conditionalMethod Dec 03 '19 at 04:24
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    It looks like $-r$ is approximately $-0.6586267543001639$. – conditionalMethod Dec 03 '19 at 04:34
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    To evaluate more precisely $f(-r)$ you will need $-r$ with more precision $-0.6586267543001639224134728305795016459409327962204365870628047777374586829997513022407599307402630816$. – conditionalMethod Dec 03 '19 at 04:51

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