The first 250 terms of an $\Bbb N$-indexed sequence in $\Bbb C$, are shown here.

What's the function $h(x):\Bbb R\to\Bbb C$ of which this is a subset, and what's $\frac{dr}{d\theta}$ as a function of either $x$ or $h(x)$ (setting $re^{i\pi\theta}=h(x)$)?

The sequence $S_n$ is given by the map $S_n=h(x):x\in2\Bbb N-1$, where $h$ is defined:

$f:2\Bbb N-1\to\frac12+\Bbb Z[\frac12]/\frac12\Bbb Z$

$f(x)=\frac x{2^p}\in\left[\frac12,1\right)$

$h(x)=x\cdot \exp{(4\pi i\cdot f(x))}$

The function's effectively a sequence of segments glued together at $x=1, \Im(h(x))=0$.

It seems obvious there must be a fairly simple continuous function $h^\times:\Bbb R\to\Bbb C$ of which this is a subset (give or take possible singularities at $0,1$). What is it? Obviously there are infinitely many, but there should be an obvious choice - I speculate the rule that chooses the canonical extension would be something like there will be an $n^{th}$ derivative of $r$ with respect to $\theta$ that's fixed or monotonic at any given point on the line.

What's the function $h^\times(x):\Bbb R\to\Bbb C$, and what's $\frac{dr}{d\theta}$ as a function of either $x$ or $h^\times(x)$?

It may be worth pointing out that $f(x) \mapsto \theta(h(x))$ sends (at least a subset of) the Prufer 2-group in its dyadic form to its 2nd roots of unity form.


The function

$$h(x) = 2^x\exp\left(2\pi i \cdot2^{x-\lfloor x\rfloor}\right)$$

is continuous, relatively simple, and obviously fits the sequence at $x = \log_2(m)$. Since $|h(x)| = 2^x$ and $\mathrm{Arg}[h(x)] = 2\pi 2^{x-\lfloor x \rfloor}$,

$$ \frac{dr}{d\theta} = \frac{dr/dx}{d\theta/dx} = \frac{2^x\ln 2}{2\pi 2^{x-\lfloor x \rfloor} \ln 2} = \frac{2^{\lfloor x \rfloor}}{2\pi} $$

As an aside, this function seems to be an approximation of the logarithmic spiral $$ z(x) = 2^x\exp(2\pi i x), $$ which has $$ \frac{dr}{d\theta} = \frac{2^x}{2\pi}. $$

  • $\begingroup$ Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly? $\endgroup$ – samerivertwice Jan 6 at 17:01
  • $\begingroup$ Do I understand correctly; the crucial part of this is that $2^{x}$ is even for whole numbers $x$ so $2^{x-\lfloor x\rfloor}\cong2^{-\lfloor x\rfloor}$ - is that right? $\endgroup$ – samerivertwice Jan 6 at 17:06
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    $\begingroup$ @user334732 Yeah, you got the translation wrong. $h(x)$ is parameterized so that it matches $S(m)$ when $x = \log_2(m)$. So you should have $(m \cos[2\pi \log_2(m)],m \sin[2\pi \log_2(m)])$ if you want $z(x)$ to look like your function. $\endgroup$ – eyeballfrog Jan 6 at 17:20
  • $\begingroup$ thanks. I was sure it would be me! I'll look again at that. $\endgroup$ – samerivertwice Jan 6 at 17:31
  • $\begingroup$ Right, thanks to your help I have simply $z(x)=x\cdot x^{(2\pi i/\log 2)}$ which I think preserves all the properties I want, and has some added properties w.r.t. continuity etc. Does that look right? $\endgroup$ – samerivertwice Jan 6 at 23:02

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