15
$\begingroup$

Let $z$ be a complex number with $|z|<1$. For every subset $A\subset\mathbb N$, the series $\sum_{m\in A}z^m$ is convergent. Denote $S(A)\in\mathbb{C}$ its sum and $\Sigma_z$ the set of all numbers $S(A)$. Remark that the cardinal of $\Sigma_z$ is (likely) that of ${\cal P}({\mathbb N})$, the continuum.

Is it possible that $\Sigma_z$ be a neighbourhood of the origin ?

$\endgroup$
  • 7
    $\begingroup$ The question is: Does there exist $|z|<1$ such that $\Sigma_z=\{\sum\limits_{n\in A} z^n: A\subseteq \mathbb N_0\}$ is a neighbourhood of $0$. $\endgroup$ – Jochen Wengenroth May 27 at 8:00
  • 2
    $\begingroup$ @JochenWengenroth Why is $\alpha\notin\mathbb Q$ necessary when $r<1$? $\endgroup$ – Emil Jeřábek May 27 at 9:14
  • 2
    $\begingroup$ To record what I just understood from @JochenWengenroth's comment: $r \geq 1/2$ is necessary because $\Sigma = z\Sigma \cup (1 + z\Sigma)$ and therefore $\mu(\Sigma) \leq 2r\mu(\Sigma)$ where $\mu$ is Lebesgue measure. ($\Sigma$ is evidently compact and therefore of finite measure.) $\endgroup$ – François G. Dorais May 27 at 9:28
  • 3
    $\begingroup$ I agree with @EmilJeřábek. Wouldn't $z = i/\sqrt{2}$ work knowing that every integer has a negabinary representation? mathworld.wolfram.com/Negabinary.html $\endgroup$ – François G. Dorais May 27 at 9:43
  • 2
    $\begingroup$ @FrançoisG.Dorais $\mu(z\Sigma) = r^2 \mu(\Sigma)$, right? So that gives $r \geq 1/\sqrt 2$. $\endgroup$ – Bart Michels May 27 at 12:18
12
$\begingroup$

The number $z = i/\sqrt2$ seems to work!

Given $x \in [-2/3,4/3]$ we can find a "negabinary" expansion $$x = \sum_{k=0}^\infty (-1)^k\frac{b_k}{2^k},$$ where each $b_k \in \{0,1\}$. Similarly, given $y \in [-2/3\sqrt2,4/3\sqrt2]$ we can find $$y = \frac{1}{\sqrt2}\sum_{j=0}^\infty (-1)^j\frac{c_j}{2^j},$$ where each $c_j \in \{0,1\}$. Therefore, $x + iy = \sum_{n \in A} z^n$ where $$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$


As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $|z| \geq 1/\sqrt2$ is necessary. By definition, $$\Sigma_z = z\Sigma_z\cup(1+z\Sigma_z).$$ If $\mu$ denotes Lebesgue measure, then $\mu(z\Sigma_z) = |z|^2\mu(\Sigma_z)$ thus $\mu(\Sigma_z) \leq 2|z|^2\mu(\Sigma_z)$. Since $\Sigma_z$ is compact, it has finite measure and thus if $\mu(\Sigma_z)>0$ then we must have $|z|^2 \geq 1/2$.

It remains open whether $|z|\geq1/\sqrt2$ and $z \notin \mathbb{R}$ is sufficient for $\Sigma_z$ to contain $0$ in its interior.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm a bit tired so the interval bounds might be a bit off... Please check my arithmetic! $\endgroup$ – François G. Dorais May 27 at 10:10
  • $\begingroup$ Denis: thank you for correcting my accidental abuse of the letter $i$. For what its worth, I think the case $z= i/\sqrt2$ is anecdotal and doesn't fully answer your question. $\endgroup$ – François G. Dorais May 27 at 11:02
  • $\begingroup$ François, it answers my question. Of course, it raises the more general question of which z has this property ?. By the way, how do you prove that $\Sigma_z$ is measurable ? $\endgroup$ – Denis Serre May 27 at 13:38
  • 5
    $\begingroup$ Denis, $\Sigma_z$ is compact since it is a continuous image of $2^{\mathbb N}$. Therefore it is measurable and has finite measure. $\endgroup$ – François G. Dorais May 27 at 16:05
  • $\begingroup$ See Davis, Chandler; Knuth, Donald E. Number representations and dragon curves-I. J. Recreational Math. 3 (1970), no. 2, 66–81. and part II in J. Recreational Math. 3 (1970), no. 3, 133–149. $\endgroup$ – Gerald Edgar May 27 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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