1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series - derivation

  1. May 21, 2009 #1
    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


    User Avatar
    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


    User Avatar

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook