Polar Square - please help!

Hello everyone!

I have recently become obsessed with the idea of a polar equation whose graph is a square.

There is an easy way to solve this problem, and it looks like this:

\(r=\sqrt{min(|\tan \theta|,|\cot \theta|)^{2} + 1} \)

That equation is really nothing more than the good old Pythagorean theorem, with r as the hypotenuse of a right triangle. One leg of the right triangle will always have length 1, and the other will either be tanθ|\tan \theta| or cotθ|\cot \theta|, whichever one is smaller.

But the real question is, is it possible to make an equation WITHOUT the min function, and without making the function piecewise, both of which feel sort of like cheating, whose polar graph is still a square? The closest thing I've found is this monster:

r=12+1cos(2θ)45+2r=-\frac{1}{\sqrt{2}+1}\left|\cos\left(2\theta\right)^{\frac{4}{5}}\right|+\sqrt{2}

Which looks like a square with the edges bent slightly out of shape. This equation is, essentially, a very distended four-petal polar rose. I have also discovered that this equation:

r=cos(nθ2)+(n2)2r=-\left|\cos\left(\frac{n\theta}{2}\right)\right|+\left(\frac{n}{2}\right)^{2}

produces a graph of a similarly slightly-squished version of a regular polygon with n sides.

So my question is, given the equations above, both of which graph as something that is ALMOST a square, is there a way to tweak them so that they look EVEN MORE like a square? Or is this the closest they get? Please help!

#Algebra

Note by Nicholas Cox
1 year ago

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Comments

r=cos(πn)cos(πnθ+2πnnθ2π)r=\frac{cos(\frac{π}{n})}{cos(\frac{π}{n}-θ+\frac{2π}{n}⌊\frac{nθ}{2π}⌋)}

this equation will accurately make any nn sided regular polygon if you want to rotate it by an angle tt then replace θθ by θtθ-t

Zakir Husain - 1 year ago

You can use Fourier series to tweak a parametric polar equation as close as you like to the shape of a square (or other polygon). See Fourier Construction of Regular Polygons https://demonstrations.wolfram.com/FourierConstructionOfRegularPolygonsAndStarPolygons/

Justin Travers - 1 year ago

thanks!

Nicholas Cox - 1 year ago

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