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Orthonormality of Fourier functions

  1. Feb 20, 2010 #1
    Looking at a Fourier expansion,

    [tex]f(x) = a_0 + a_1 \cos{x} + b_1 \sin{x} + a_2 \cos{2 x} + b_2 \sin{2 x} + ...[/tex],

    I would expect the coefficients to be the projections of [tex]f(x)[/tex] on the different Fourier functions, which should form an orthonormal basis. This doesn't seem to be the case, though. The formulae for the coefficients say

    [tex]a_0 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) 1 dx[/tex];

    [tex]a_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) 2 \cos{n x} dx[/tex];

    [tex]b_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) 2 \sin{n x} dx[/tex].

    Now, assuming that the inner product is defined as

    [tex]<f,g> = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) g(x) dx[/tex],

    these coefficients are not simply the projections, since the sines and cosines get a factor two. Further more, these sines and cosines do not form an orthonormal set. So what exactly is going on? Are my assumptions about the nature of the coefficients or about the expected orthonormality of the functions wrong?

    I also noticed that if you use the exponential form of the Fourier series, all the functions are perfectly orthonormal and the coefficients are indeed the projections on these functions. Perhaps something happens when changing from exponential form to sine and cosine form?
  2. jcsd
  3. Feb 20, 2010 #2
    So Redefine the inner product to be:


    And then the sines and cosines will be orthonormal to each other ( the constant function will need refactoring, it's the only function that doesn't blend nicely into the set).

    Convention gives the next sum:

    [tex]f(x) -> \frac{a_{0}}{2}+\sum^{\infty}_{n=1}a_{n}cos(nx)+b_{n}sin(nx)[/tex]



    I must add that dealing with orthonormallity is quite artificial. You can always normalize every function given an inner product (assuming of course, it is of the integral form). The center of the Fourier Series (and Sturm-Liouville comes in too) Theorem is the orthogonality.

    P.S. About your last paragraph regarding the exponential basis. Yes, there is this 1/2 factor when considering Euler's Identities:


    Perhaps this is what causes the orthonormality of the exponentials with the 1/(2pi) inner product.
  4. Feb 20, 2010 #3
    I hadn't though of it. You can of course absorb a factor 2 into [tex]a_0[/tex] to make your functions orthonormal. Thanks.
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