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Trigonometric interpolation polynomial

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]t_j=j/100, a_j=j, b_j=-j[/itex], for j=0,1,...,99. Define [tex]f(t)=\sum\limits_{k=0}^{99} (a_k\cos(2\pi kt)+b_k\sin(2\pi kt))[/tex]
    Determine the values of [itex]c_l, d_m[/itex] for l= 0,...5, m=1,...,4, so that [tex]P(t)=\frac{c_0}{2}+\sum\limits_{k=1}^4 (c_k\cos(2\pi kt)+d_k\sin(2\pi kt))+c_5\cos(10\pi kt)[/tex]
    is the least squares approximation to the data point [itex](t_j,f(t_j))[/itex] for j=0,...,99.


    2. Relevant equations
    [tex] c_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\cos(2\pi kt_j)[/tex]
    [tex] d_k=\frac{1}{50}\sum\limits_{j=0}^{99} f(t_j)\sin(2\pi kt_j)[/tex]


    3. The attempt at a solution
    It's clear that I have to evaluate [itex]f(t_j)[/itex] first, but I don't know how to. I've tried simplifying the expression for [itex]f(t_j)[/itex] a bit but this is all I can get
    [tex]f(t_j)=100\sum\limits_{k=1}^{49}\cos(2\pi kt_j)+50\cos(2\pi50t_j)+\sum\limits_{k=1}^{49}(100-2k)\sin(2\pi kt_j))[/tex]
    since [itex]\cos(2\pi(100-k)t_j)=\cos(2\pi kt_j), \sin(2\pi(100-k)t_j)=-\sin(2\pi kt_j), \sin(2\pi 50t_j)=0.[/itex]
    I would be much appreciated if someone could help me evaluate this summation, thanks!
     
  2. jcsd
  3. Dec 21, 2013 #2

    vela

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    You have
    $$f(t) = \sum_{k=0}^{99} [k \cos(2\pi k t) - k \sin(2\pi k t)].$$ You could try using the fact that
    $$k \cos(2\pi k t) - k \sin(2\pi k t) = \frac{1}{2\pi}\frac{\partial}{\partial t} [\sin(2\pi k t) + \cos (2\pi k t)].$$ You can evaluate the sums of sin and cos by considering the geometric series ##(e^{i2\pi t})^k##.
     
  4. Dec 22, 2013 #3
    Thanks for the idea, but I can't seem to arrive at anything useful. Here's what I have done
    $$f(t) = \sum_{k=0}^{99} [k \cos(2\pi k t) - k \sin(2\pi k t)]=\frac{1}{2\pi}\frac{\partial}{\partial t}\sum_{k=0}^{99} [\sin(2\pi k t) + \cos (2\pi k t)].$$
    $$\sum_{k=0}^{99}\sin(2\pi k t)=\mathrm{Im}\sum_{k=0}^{99} (e^{i2\pi t})^k=\ldots=\frac{\sin(100\pi t)\sin(99\pi t)}{\sin (\pi t)}$$
    Similarly, $$\sum_{k=0}^{99}\sin(2\pi k t)=1+\frac{\cos(100\pi t)\sin(99\pi t)}{\sin (\pi t)}$$
    I'm not gonna differentiate this whole thing wrt t, am I?
     
  5. Dec 22, 2013 #4

    vela

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    Yeah, that does look like it's going to be tedious, doesn't it? Perhaps you can differentiate a little earlier when you have the results of the sums in terms of complex exponentials still.
     
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