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 2020-05-02, 10:20 #1 jnml   Feb 2012 Prague, Czech Republ 2×34 Posts Divisibility function $$\forall N \ne 0,\ x \in [1,\ N]: \ sin \, x \pi +\imath \, sin \frac{N \pi}{x} = 0 \leftrightarrow x|N.$$ I'm not aware of any practical value, but I like the pictures, produced in Maxima by draw2d( nticks=5000, grid=true, xrange=[-1,1], yrange=[-1,1], terminal='pngcairo, parametric(sin(%pi*x),sin(%pi*n/x),x,1, n) ), n=; edit: fixed tex formatting Attached Thumbnails         Last fiddled with by jnml on 2020-05-02 at 10:30
 2020-05-02, 10:22 #2 jnml   Feb 2012 Prague, Czech Republ 2×34 Posts 6 to 9 Attached Thumbnails
 2020-05-02, 10:23 #3 jnml   Feb 2012 Prague, Czech Republ 101000102 Posts 10 to 13 Attached Thumbnails
 2020-05-02, 10:25 #4 jnml   Feb 2012 Prague, Czech Republ 2·34 Posts 14 to 17 and that's all. Attached Thumbnails
 2020-05-02, 11:52 #5 Nick     Dec 2012 The Netherlands 142510 Posts Look up Lissajous curves.
2020-05-02, 12:37   #6
jnml

Feb 2012
Prague, Czech Republ

2×34 Posts

Quote:
 Originally Posted by Nick Look up Lissajous curves.
I don't think the function in #1 qualifies as a Lissajous curve.
The reason I see is that the argument of the sinus function
of the complex part is not a proportional to $$t$$, but to its inverse.
In fact, that's what makes the sum a divisibility function, which
the Lissajous curve - AFAICT - is not.

2020-05-02, 12:41   #7
Dr Sardonicus

Feb 2017
Nowhere

335610 Posts

Quote:
 Originally Posted by Nick Look up Lissajous curves.
Related, but Lissajous figures are parameterized by x = sin(L1(t)), y = sin(L2(t)) (or cosine) where L1 and L2 are linear functions.

The parameterization here is of the form x = sin(L1(t)), y = sin(1/L2(t)), in a range of t-values that insures L2(t) is not 0.

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