# Of which continuous line in $\Bbb R\to\Bbb C$ is this sequence a subset?

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 $$\frac{dr}{d\theta} = \frac{2^x}{2\pi}.$$

• Is this your function $z(x)$?: desmos.com/calculator/r8xi2zvogc or have I translated it wrongly? – samerivertwice Jan 6 at 17:01
• 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? – samerivertwice Jan 6 at 17:06
• @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. – eyeballfrog Jan 6 at 17:20
• thanks. I was sure it would be me! I'll look again at that. – samerivertwice Jan 6 at 17:31
• 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? – samerivertwice Jan 6 at 23:02