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Old 2020-05-02, 10:20   #1
jnml
 
Feb 2012
Prague, Czech Republ

22·41 Posts
Default 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=<insert N here>;


edit: fixed tex formatting
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Last fiddled with by jnml on 2020-05-02 at 10:30
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Old 2020-05-02, 10:22   #2
jnml
 
Feb 2012
Prague, Czech Republ

22·41 Posts
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6 to 9
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Old 2020-05-02, 10:23   #3
jnml
 
Feb 2012
Prague, Czech Republ

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10 to 13
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Old 2020-05-02, 10:25   #4
jnml
 
Feb 2012
Prague, Czech Republ

101001002 Posts
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14 to 17 and that's all.
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Old 2020-05-02, 11:52   #5
Nick
 
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Dec 2012
The Netherlands

22·3·53 Posts
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Look up Lissajous curves.
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Old 2020-05-02, 12:37   #6
jnml
 
Feb 2012
Prague, Czech Republ

2448 Posts
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Quote:
Originally Posted by Nick View Post
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.
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Old 2020-05-02, 12:41   #7
Dr Sardonicus
 
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Feb 2017
Nowhere

380210 Posts
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Quote:
Originally Posted by Nick View Post
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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