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Trigonometric Orthogonality Query

  1. Oct 18, 2013 #1
    Hello everyone, I've wandered PF a few times in the past but never thought I'd join, here I am, how exciting.

    To keep it short I'm trying to understand the proof behind Fourier Series and can't quite get to grips with basic trigonometric orthogonality.
    I understand that sin and cos are naturally orthogonal but I'm having difficulty understanding this:

    \int_{-\pi}^{\pi}\cos(mx)\cos(nx)\, dx=\pi\delta_{mn} \,\,\,where\,\delta_{mn}=\begin{cases}
    &\text{1 if } m=n \\
    &\text{0 if } m\neq n

    It makes sense that the area is [itex]\pi[/itex] if [itex]m=n[/itex] because you get this waveform and zero if m is an integer multiple of n because you get this waveform (here the negative areas cancel the positive areas from [itex]-\pi[/itex] to [itex]\pi[/itex])

    BUT... If [itex]m\neq n[/itex] AND is sufficiently close e.g (m=2, n=2.1) you get this waveform, in which case the area from [itex]-\pi[/itex] to [itex]\pi[/itex] is clearly not zero.

    What gives? Does this only apply if m and n are different by integer multiples?
    Thanks for your time! (:
    Last edited: Oct 18, 2013
  2. jcsd
  3. Oct 18, 2013 #2
    For both m,n ∈ ℕ, we have the following proof for this orthogonality relation:

    Edit : posted it as an attachment because it wouldn't work otherwise.

    Attached Files:

    Last edited: Oct 18, 2013
  4. Oct 18, 2013 #3
    This a plot visualization of the area when m and n are different, from -π to π. In this case, n=2 and m=3.

    Can you see the area is equal to 0?
  5. Oct 18, 2013 #4
    I just realized what you did wrong...

    Your waveform has (2.1x) and not (2,1x) in the second cosine.

    Try changing 2.1x to 2,1x.

    Many computing softwares read the input 2.1 as 2*1.

    With 2,1x, you'll get the expected result. ;)
  6. Oct 18, 2013 #5
    The software produces the correct waveforms, it actually glitches when using commas (try cos(2,0x)).
    The waveform is what I expected, but it just doesn't match up with the mathematics which states it should equal zero.

    It only equals zero if you take the limit from [itex]-\infty[/itex] to [itex]\infty[/itex]
  7. Oct 18, 2013 #6
    The integral does not converge if you take the limit from -∞ to ∞. It only converges (to 0) from -nπ to +nπ where n is an integer. The attached visual representation of the integral in my earlier post shows how the area is equal to 0.
    Last edited: Oct 19, 2013
  8. Oct 18, 2013 #7
    Yes, m and n must both be integers. Otherwise, as you pointed out, the two functions are not orthogonal. So, not only must m and n but integer differences of each other, they must both be integers.
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