21
$\begingroup$

The plots of $\sin(kx)$ over the real line are somehow boring and look essentially all the same:

enter image description here

For larger $k$ you cannot easily tell which $k$ it is (not only due to Moiré effects):

enter image description here

But when plotting $\sin(kx)$ over the unit circle by

$$x(t) = \cos(t) (1 + \sin(kt))$$ $$y(t) = \sin(t) (1 + \sin(kt))$$

interesting patterns emerge, e.g. for $k = 1,2,\dots,8$

enter image description here


Interlude: Note that these plots are the stream plots of the complex functions

$$f_k(z)=\frac{1}{2i}(z^k - \overline{z^k})z $$

on the unit circle (if I didn't make a mistake). Note that $f_k(z)$ is not a holomorphic function.

You may compare this with the stream plot of

$$g_k(z)=\frac{1}{2i}(z^k - \overline{z^k}) = f_k (z)/z$$

with $g_k(e^{i\varphi}) = \sin(k\varphi) $:

enter image description here

[End of the interlude.]


Even for larger $k$ one still could tell $k$:

enter image description here

Furthermore you can see specific effects of rational frequencies $k$ which are invisible in the linear plots. Here are the plots for $k=\frac{2n +1}{4}$ with $n = 1,2,\dots,8$:

enter image description here

The main advantage of the linear plot of $\sin(kx)$ is that it has a simple geometrical interpretation resp. construction: It's the plot of the y-coordinate of a point which rotates with constant speed $k$ on the fixed unit circle:

enter image description here

Alternatively, you can look at the sine as the projection of a helix seen from the side. This was the idea behind one of the earliest depictions of the sine found at Dürer:

enter image description here

Compare this to the cases of cycloids and epicycles. These also have a simple geometrical interpretation - being the plots of the x- and y-coordinates of a point on a circle that rolls on the line

enter image description here

resp. moves on another circle with constant speed

enter image description here

My question is:

By which geometrical interpretation resp. construction (involving circles or ellipses or whatsoever) can the polar plots of $\sin$ be seen resp. generated? Which construction relates to the construction of $\sin$ by a rotating point on a circle in the way that the construction of epicycles relates to the construction of cycloids?


Just musing: Might this question have to do with this other question on Hidden patterns in $\sin(kx^2)$? (Probably not because you cannot sensibly plot $\sin(kx^2)$ radially, since there is no well-defined period.)

$\endgroup$
  • $\begingroup$ Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it. $\endgroup$ – timtfj Jan 30 at 11:08
  • $\begingroup$ (I still upvoted though) $\endgroup$ – timtfj Jan 30 at 11:16
  • 1
    $\begingroup$ "Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation). $\endgroup$ – Hans-Peter Stricker Jan 30 at 11:22
  • $\begingroup$ @HansStricker I think it's more common to use "re" instead of "resp". $\endgroup$ – Jam Jan 30 at 12:23
  • $\begingroup$ @HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :) $\endgroup$ – Jam Jan 30 at 12:31
2
$\begingroup$

Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:

enter image description here

– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:

enter image description here

– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:

The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation around one of their points.


The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $\sin(e^{ik\varphi})$ not over the circle (as the base line), but from the origin. Here for $0 \leq \varphi < \pi$ (left) and $\pi < \varphi < 2\pi$ (right):

enter image description hereenter image description here

$\endgroup$
  • $\begingroup$ better you take just a ray from the origin: then it is easier to controll the parameters $\endgroup$ – G Cab Jan 30 at 14:58
  • $\begingroup$ @GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.) $\endgroup$ – Hans-Peter Stricker Jan 30 at 15:01
  • $\begingroup$ judging by eye from the animation, that's constructed by crossing the entire line ($\pm \rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes. $\endgroup$ – G Cab Jan 30 at 15:50
1
$\begingroup$

I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.

Lissajous_animation

$\endgroup$
  • $\begingroup$ No, the question is not related to the Lissajous curves. $\endgroup$ – Yves Daoust Jan 30 at 13:58
  • $\begingroup$ @YvesDaoust: now that the question has been edited, it is more clear. $\endgroup$ – G Cab Jan 30 at 15:00

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.