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Fourier expansion can be used to represent any periodic function in one variable.

Closed surfaces in 3D can be built out of spherical harmonics.

Is there a similar expansion to represent a curve of any shape, like the following one?

enter image description here

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    $\begingroup$ Are you thinking of the curve as a function y(x) ? If so, then it's not a single valued function of x, so the normal elementary Fourier transform cannot be applied. If, however, you were thinking of functions ON the curve, then they'd be periodic by construction so you could Fourier transform them. $\endgroup$ – twistor59 Nov 18 '11 at 16:47
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    $\begingroup$ I think this may be a good candidate for migration to Mathematics... thoughts? $\endgroup$ – David Z Nov 18 '11 at 16:51
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    $\begingroup$ A classic example to this is "epicyles" used to calculate planets trajetories in prekeplerian times. AFAIK this method is still used in in practical astronomy (directing telescopes, space flight) because it is much easier to program. $\endgroup$ – Georg Nov 18 '11 at 17:06
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    $\begingroup$ @Revo, your curve is a function of $(x(t), y(t))$, just do two Fourier expansions for those two functions $\endgroup$ – lurscher Nov 18 '11 at 18:45
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    $\begingroup$ demonstration: youtube.com/watch?v=QVuU2YCwHjw $\endgroup$ – Mark Eichenlaub Nov 18 '11 at 23:53
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As lurscher suggests in a comment, in the case of a closed curve, one could consider a periodic parametrization of the curve

$${\bf f}(\theta)~=~{\bf f}(\theta+2\pi)~\in~\mathbb{R}^2, \qquad {\bf f}(\theta)~=~(x(\theta),y(\theta)). $$

Then define Fourier coefficients in the standard way

$$ {\bf c}_n({\bf f})~:=~ \int_0^{2\pi} \frac{{\rm d}\theta}{2\pi} e^{-in\theta}~{\bf f}(\theta). $$

(The Fourier coefficients ${\bf c}_n({\bf f})$ are well-defined if the coordinate functions $x,y$ are Lebesgue integrable $x,y\in{\cal L}^1(\mathbb{R}/2\pi\mathbb{Z}).$) The Fourier series for ${\bf f}$ is vector-valued

$$\sum_{n\in\mathbb{Z}}{\bf c}_n(f) ~e^{in\theta}.$$

A similar approached works also for a closed curve in higher dimensions. In the 2 dimensional case, one may identify the plane $\mathbb{R}^2\cong \mathbb{C}$ with the complex plane, as Greg P, Mark Eichenlaub, and J.M. point out.

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  • $\begingroup$ We can also think of it as just a usual complex-valued Fourier transform, since complex numbers can represent two dimensions. (I now see that Greg P pointed this out in the comments to the main question.) $\endgroup$ – Mark Eichenlaub Nov 18 '11 at 23:52
  • $\begingroup$ One can also take $\mathbf f(\theta)=x(\theta)+i\,y(\theta)$; i.e., consider a complex-valued function instead of a vector-valued function. $\endgroup$ – J. M. isn't a mathematician Nov 19 '11 at 0:03

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