Parametrizing a Closed Curve in \mathbb{R}^3

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Homework Help Overview

The discussion revolves around the parametrization of a closed curve in \(\mathbb{R}^3\). The original poster is exploring the possibility of using a Fourier series for this parametrization while ensuring the curve does not intersect itself.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster suggests using a Fourier series for parametrization and questions if this approach is valid given the boundary conditions. Some participants propose defining the curve as a function with specific properties, while others inquire about the relevance of Fourier series to the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the adequacy of proposed solutions. There is no explicit consensus on the best approach yet, but various ideas are being examined.

Contextual Notes

Participants are considering the implications of the boundary condition \(r(0) = r(1)\) and its relationship to Fourier series, as well as the requirement for the curve to be non-intersecting.

Dox
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Hello.

In a certain problem I'm interested on, I need to write a general form of the parametrization of a closed curve on [tex]\mathbb{R}^3[/tex].

I thought in parametrize it using a kind of Fourier series. Could it be possible?

Thing become even worse 'cause I'd like to the curve doesn't cross itself.

Every idea is welcome.

Best wishes.
 
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Would it not be sufficient to say that such a curve is a function [itex]r(t)=(x(t),y(t),z(t))[/itex], [itex]t\in [0,1][/itex] such that r(t1)=r(t2) iff r1=0, r2=1.
 
quasar987 said:
Would it not be sufficient to say that such a curve is a function [itex]r(t)=(x(t),y(t),z(t))[/itex], [itex]t\in [0,1][/itex] such that r(t1)=r(t2) iff r1=0, r2=1.

Thanks for your answer Quasar987, but that exactly the point. In order to satisfy the boundary condition I must expand in Fourier series, Isn't it?

Thank you very much.
 
I don't know what you mean.

Why do you consider the answer I wrote incomplete/inadequate?
 
I mean:

The condition is [tex]r(0)=r(1)[/tex].

In general a function satisfying that condition could be expand as a Fourier series, Isn't it? Because sine and cosine are periodic functions.
 
If the function is not too wild, yes. But again, what does this have to do with the problem? :confused:
 

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