You can write an equation for a circle, but why can't you write an equation for a triangle or any other polygon? By equation I mean an equation that is not just a piecewise equation of lines.
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2$\max(|x|,| y|) = 1$ isn't good enough for you? More seriously, though, part of the reason has to do with polygons being inherently piece-wise linear things. The curves that can be defined by polynomials have a more "organic" look to them. – Arthur May 07 '15 at 23:04
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2Have you heard of Schwarz-Christoffel mappings? – Erick Wong May 07 '15 at 23:10
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1If you're willing to accept 'diamonds' as squares, $y = \pm \left( 1- |x| \right)$ makes a perfectly good square (restricting $|x| \leq 1$). I'd be willing to bet that any vertical lines are very hard to create "functions" for. – pjs36 May 07 '15 at 23:12
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Polygons are made of line segments. A parametrization would encounter a discontinuous derivative moving from one segment to the next. Inherent in the construction of polygons is that they are pieces of lines, segmented and attached to one another. So you can expect a piece-wise description.
Circles neither have these discontinuous nor are they even composed of line segments.
zahbaz
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There is nothing discontinuous about polygons. In fact, it's kind of an asumption that their boundary be a simple, closed curve. – Arthur May 07 '15 at 23:10
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Being composed of line segments doesn't seem to be important beyond the non-smoothness. It's very easy to map a circular arc to a single line segment. – Erick Wong May 07 '15 at 23:13
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2The derivative would not necessarily be discontinuous. $t\mapsto (t^3,|t^3|)$ is a continuously differentiable parametrization of the graph of $|x|$. – Matt Samuel May 07 '15 at 23:29
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1A parametrization with image a polygon need not have a discontinuous derivative. Simply have the derivative at the corner be instantaneously zero. – aes May 07 '15 at 23:42
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@aes The derivative of a parametrization of a polygon will not be defined at the vertices according to the limit definition. If you want to declare by fiat that you want to define a function which is the derivative of the parametrization everywhere but the vertices but is 0 at the vertices you can do that, but that's not the same as the derivative existing at the vertices. – anon May 07 '15 at 23:45
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@MattSamuel. That example is enlightening, thanks. I'm having a tough time reconciling it with the fact that polygons are not smooth, though. Hell, I'm having a hard time intuitively reconciling it with the corner on the graph of $|x|$. – zahbaz May 07 '15 at 23:55
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3@zahbaz I think of it as smoothing out the corner by composing with a sufficiently smooth bijective function. A particle along the curve slows down and stops at the corner, then speeds back up and goes off to the right. – Matt Samuel May 07 '15 at 23:59
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They can be defined by equations. The only issue is what kind of functions you are willing to use in the equations. We can use artificial piecewise-defined functions. More "natural" would be the so-called Schwarz-Christoffel mappings in complex analysis, with some finagling I imagine.
Perhaps you want to restrict your attention to algebraic equations. If any line segment is in an algebraic set (the zero loci of some algebraic equations), then the whole line is as well. One way to see this is to write the equation(s) as $f(x)=0$, then parametrize the line and plug into $f(x)$ and finally differentiate with respect to the parameter: since the derivative is a polynomial that vanishes on an interval it must be everywhere zero. As polygons have line segments which are not complete lines, they cannot be the solution set to a system of algebraic equations.
anon
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