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http://www.wolframalpha.com/input/?i=y%5E2+%3D+%7Ccos%28pi*x%2F2%29%7C

The generation for the infinite string of circles on $y = 0$.

Is there a relation that generates an infinite number of square adjacent packed circles on the cartesian plane?

How about hexagonally adjacent packing?

Peter Phipps
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Jack Tiger Lam
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1 Answers1

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There isn't one single relation that will create square or hexagonal packings. However, if you allow piecewise functions, you can easily make "functions".

I believe that the function given above does not give circles, as the Julián Aguirre mentioned. This is because the slope of the above function never approaches infinity (it peaks at $\frac{\pi}{2}$), whereas a function that would output circles would be vertical.

  • I remember an equation that generated five circles arranged in the side of a die. It was a nested trig equation of some kind... And it was centred on the origin. – Jack Tiger Lam May 25 '15 at 12:27
  • Could it have been parametrically defined? For sure, this particular equation does not give a circle. It's also possible that the circles were close but not perfectly a circle. –  May 25 '15 at 12:55
  • It was parametrically defined. There might have been a mod involved. – Jack Tiger Lam May 26 '15 at 01:01
  • The parametric relationship for a circle is $x(t)=\cos(t)$ and $y(t) = \sin(t)$. In this case, $x^2 = \cos^2(t)$ and $y^2 = \sin^2(t)$. But we know that $\sin^2(\theta)+\cos^2(\theta) = 1$. So $x^2+y^2 = 1$ is the equation of a circle. This does not, however, mean that you can make a circle out of sin() equations with y and x. –  May 26 '15 at 08:44