Fourier-like expansion of a closed curve in 2D

Question

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?

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo

Comments

@Revo, sure, which are better known as just do two Fourier expansions for those two functions

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user lurscher

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You could think of the path of the curve as being in the complex plane. Then you can do the Fourier transform of that complex function.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Greg P

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@Revo (4 comments up): having physics applications is not a sufficient condition to be on topic on Physics.SE. I'm going ahead and migrating this.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user David Z

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demonstration: youtube.com/watch?v=QVuU2YCwHjw

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Mark Eichenlaub

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Since Georg brought up epicycles: have a look at this.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Ｊ. Ｍ.

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@Revo, your curve is a function of $(x(t), y(t))$, just do two Fourier expansions for those two functions

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user lurscher

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@DavidZaslavsky Why it should be moved to mathematics? Some shape in space can be built out of spherical harmonics which has so many applications, similarly the concepts of multipole moments and expansion which all have physics applications. If we can expand closed surface, why cannot we expand a closed line?

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo

Answer 1

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.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Qmechanic

Comments

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.)

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Mark Eichenlaub

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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.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Ｊ. Ｍ.