Trigonometric Orthogonality Query

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SUMMARY

The discussion centers on the proof of trigonometric orthogonality in Fourier Series, specifically the integral relation \(\int_{-\pi}^{\pi}\cos(mx)\cos(nx)\, dx=\pi\delta_{mn}\). It is established that this relation holds true only when both \(m\) and \(n\) are integers, and the integral converges to zero only when \(m\) and \(n\) differ by integer multiples. The confusion arises when non-integer values are mistakenly used, leading to incorrect interpretations of the waveform and integral results.

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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:

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

It makes sense that the area is \pi if m=n 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 -\pi to \pi)

BUT... If m\neq n AND is sufficiently close e.g (m=2, n=2.1) you get this waveform, in which case the area from -\pi to \pi 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:
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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.
 

Attachments

  • png-1.latex.png
    png-1.latex.png
    5.2 KB · Views: 490
Last edited:
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?
 

Attachments

  • WolframAlpha--integral_cos_2x__cos_3x__from__pi_to_pi__Visual_representation_of_the_integral__20.gif
    WolframAlpha--integral_cos_2x__cos_3x__from__pi_to_pi__Visual_representation_of_the_integral__20.gif
    3.7 KB · Views: 512
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. ;)
 
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 -\infty to \infty
 
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:
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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