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I came across this interesting blog post that posits a curious hypothesis and I would like to share my understanding of it here.

First, my understanding is that for complex $a$, $1^a$ is multi-valued, given by $$1^a=e^{2i \pi n a }, n\in \mathbb{Z}$$

and for irrational $a$ this leads to countably infinite values of the $a$th power of unity.

Consider as a special case the $\pi$th roots of unity ($a=1/\pi$):

$$1^{1/\pi}=e^{2i n}=\cos 2n +i\sin 2n, $$

and for positive integer $N$, consider the set

$$P_N\equiv \bigcup_{n=0}^{N-1}\{\cos 2n+i\sin 2n\}.$$

An interesting pattern appears to emerge when one plots $P_N$ on the unit circle for various $N$:

$P_3:\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad P_4$:

enter image description here enter image description here

$P_{22}:\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad P_{23}$:

enter image description hereenter image description here

$P_{355}:\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad P_{350}$:

enter image description hereenter image description here

Note how the points in $P_N$ appear to be equally spaced on the unit circle for certain special $N$ (e.g. $N=3,22,355$) but form clusters for other $N$ (highlighted are clear departures from equal spacing).

The significance of the special $N$ is that they are numerators of the convergents obtained from the continued fraction representation of $\pi$:

$$\pi =[3;7,15,1,292,...],\\ [3]=\frac{3}{1},[3;7]=\frac{22}{7},[3;7,15]=\frac{333}{106},[3;7,15,1]=\frac{355}{113}.$$

It seems not all convergents correspond with equal spacing:

$P_{333}$:

enter image description here

So to summarize my questions:

  1. If the points in $P_N$ are almost* equally spaced, is $N$ a numerator of a convergent of $\pi$? Or is this simply an embarrassing case of mathematical apophenia?
  2. If 1. is true, why? And why do some convergents not correspond with equal spacing?

You can obtain the plots above by this Desmos calculator (for large $N$, you may have to zoom in quite a bit to see unequal spacing).


*Update: As the comments point out, they are not exactly equally spaced. For three points, equal spacing would correspond with angles $0,2\pi/3,4\pi/3,$ but these are rather close to $0,2,4$ (i.e. $2n$ for $n=0,1,2$). Is there a justification behind these approximants, somewhat analogous to almost integers?

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    $\begingroup$ First of all, what a nice post! Secondly, The main question boils down to the points not actually being equally spaced, just very-very close to equally spaced in some cases. For the case of $n=3$, you can almost see this on the drawing, but to convince yourself, calculate the distance between the complex values of $\cos(2n)+i\sin(2n)$, $n=0,1,2$, which will all look something like $\sqrt{cos^2(2k)+sin^2(2l)}, k,l \in \{0,1,2\}, k \ne l$. They're not exactly the same. $\endgroup$
    – Daniel P
    Commented Mar 4, 2022 at 0:48
  • $\begingroup$ @DanielP Yes, good point. I guess it is analogous to "almost integers" then. $\endgroup$ Commented Mar 4, 2022 at 0:53
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    $\begingroup$ Well, I'd say more like "almost equal to a multiple of $\pi$". Again for the case of $n=3$, instead of stepping $2/3$'s along the circle, which would be a distance of $4\pi/3 $, considering the length of the curve of the unit circle is $2\pi$, we make a step of size $4$ (which is pretty close to $4\pi/3 $). $\endgroup$
    – Daniel P
    Commented Mar 4, 2022 at 1:00
  • $\begingroup$ I might have answered the question in a comment... Well, someone can reap the benefits of posting an actual, well formatted answer. $\endgroup$
    – Daniel P
    Commented Mar 4, 2022 at 1:02
  • $\begingroup$ @DanielP Thanks, updated the question:) $\endgroup$ Commented Mar 4, 2022 at 1:03

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Writing this answer with the consent of Sir Daniel, as mentioned in the comments.

The main question boils down to the points not actually being equally spaced, just very-very close to equally spaced in some cases. For the case of $n=3$, you can almost see this on the drawing, but to convince yourself, calculate the distance between the complex values of $\cos(2n)+i\sin(2n)$, $n=0,1,2$ which will all look something like $\sqrt{\cos^2(2k)+\sin^2(2l)}, k,l \in \{0,1,2\}, k \ne l$

They're not exactly the same.

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