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Closed curve on R^3

  1. Jun 5, 2007 #1

    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.
     
  2. jcsd
  3. Jun 5, 2007 #2

    quasar987

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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.
     
  4. Jun 5, 2007 #3

    Dox

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    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.
     
  5. Jun 5, 2007 #4

    quasar987

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    I don't know what you mean.

    Why do you consider the answer I wrote incomplete/inadequate?
     
  6. Jun 5, 2007 #5

    Dox

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    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.
     
  7. Jun 5, 2007 #6

    quasar987

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    If the function is not too wild, yes. But again, what does this have to do with the problem? :confused:
     
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