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  Fourier-like expansion of a closed curve in 2D

+ 3 like - 0 dislike
72 views

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

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Revo
asked Nov 18, 2011 in Mathematics by Revo (260 points) [ no revision ]
Most voted comments show all comments
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.

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user twistor59
I think this may be a good candidate for migration to Mathematics... thoughts?

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

This post imported from StackExchange Mathematics at 2014-06-02 11:01 (UCT), posted by SE-user Georg
@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
@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
Most recent comments show all comments
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
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.

1 Answer

+ 4 like - 0 dislike

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
answered Nov 18, 2011 by Qmechanic (2,860 points) [ no revision ]
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
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.

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