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Fourier series - derivation

  1. May 21, 2009 #1
    Hello,
    everywhere I can see this
    [tex]a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \cos(nt)\, dt[/tex]
    [tex]b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) \sin(nt)\, dt[/tex]
    etc... I can't find, how to derive this formulas. I'm really tired and a bit confused of this formulas, because I can't find possible way to derive them. I don't like only formula application, but I want to know, what is that formula about.
    Thank you...
     
  2. jcsd
  3. May 21, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you have an "inner product space", that is, an vector space with an inner product defined on it, together with an orthonormal basis, [itex]v_1, v_2, ....[/itex], that is such that [itex]<v_i, v_j>= 0[/itex] if [itex]i\ne j[/itex] and [itex]<v_i, v_i>= 1[/itex] for all i, and want to write v as a linear combination, [itex]v= a_1v_1+ a_2v_2+ ...+ a_nv_n[/itex], then [math]a_i= <v, v_i[/itex]. What you have is a vector space with basis [itex]cos(nx)[/itex], [itex]sin(nx)[/itex] with inner product [itex]<f, g>= \frac{1}{\pi}\int_{-\pi}^\pi f(t)g(t)dt[/itex] which leads to the given formulas.
     
  4. May 22, 2009 #3

    jtbell

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    Staff: Mentor

    It works because the functions sin(nt) for different values of n are orthogonal to each other, that is,

    [tex]\int^{\pi}_{-\pi} {\sin(nt) \sin (mt) dt} = 0[/tex]

    for [itex]n \ne m[/itex], and

    [tex]\int^{\pi}_{-\pi} {\sin^2(nt) dt} = \pi[/tex]

    Likewise for cosines. Try a few examples if you like. Therefore if you have a function

    [tex]f(t) = b_1 \sin (t) + b_2 \sin (2t) + b_3 \sin (3t) + ...[/tex]

    then, for example, letting n = 2:

    [tex]\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \int^{\pi}_{-\pi} {\sin (t) \sin (2t) dt}
    + b_2 \int^{\pi}_{-\pi} {\sin^2 (2t) dt}
    + b_3 \int^{\pi}_{-\pi} {\sin(3t) \sin (2t) dt} + ...[/tex]

    [tex]\int^{\pi}_{-\pi} {f(t) \sin (2t) dt} = b_1 \cdot 0 + b_2 \cdot \pi + b_3 \cdot 0 + ...[/tex]
     
  5. May 23, 2009 #4
    Thank you. Now I understand. Thanks Thanks
     
  6. May 23, 2009 #5
    It is easier to work with the basis functions

    e_n(x) = exp(i n x)

    and define the inner product as

    <f,g> = 1/(2 pi) Integral from minus pi to pi of f(x)g*(x) dx
     
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