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- Thread starter Alec Neeson
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I think you mean ##a_n## and ##b_n##. And I guess you are talking about the Fourier coefficients. They are just numbers and it doesn't make any sense to talk about numbers "being 90 degrees apart".

Orthogonality is a generalization of "perpendicular". Two nonzero vectors in 3D are perpendicular if their dot product is zero:$$

\langle a_1,a_2,a_3\rangle \cdot \langle b_1,b_2,b_3\rangle = a_1b_1 + a_2b_2 + a_3b_3 = 0$$This idea is generalized to functions by the definition: Functions ##f## and ##g## are orthogonal with respect to a weight function ##w(t)>0## on an interval ##[a,b]## if ##\int_a^b f(t)g(t)w(t)~dt = 0##. Often in classical Fourier series, ##w(t) = 1## Here the sum in the dot product corresponds to the integral of the two functions.

The fact that the functions ##\{\sin(\frac{n\pi x}{p}),\cos(\frac{n\pi x}{p})\}## are orthogonal on ##[-p,p]## is what allows you to get nice closed formulas for the coefficients ##a_n## and ##b_n## in Fourier series. Any text on FS will explain this in detail.

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ElijahRockers

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Remember how you used {i, j, k} to represent orthogonal unit vectors? It's the same idea with {cos(x), cos(2x), ... , cos(Nx)}.

The same way you could represent any 3D vector as xi + yj + zk, you can represent any function as a sum of sines and cosines of varying frequency.

cos(1x) is orthogonal to cos(.9999x) -> meaning the slightest variation in frequency will result in a pair of orthogonal functions...

I think. Anyone in the know, please feel free to correct or confirm my suspicions.

EDIT: Cut out some stuff I wrote that was even confusing to me.

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If that's a little wordy for you, maybe I can dumb it down. Just keep in mind I am only now learning this stuff myself.

Remember how you used {i, j, k} to represent orthogonal unit vectors? It's the same idea with {cos(x), cos(2x), ... , cos(Nx)}.

The same way you could represent any 3D vector as xi + yj + zk, you can represent any function as a sum of sines and cosines of varying frequency.

cos(1x) is orthogonal to cos(.9999x) -> meaning the slightest variation in frequency will result in a pair of orthogonal functions...

I think.

No. That isn't true. Orthogonality depends very much on the particular frequencies and the interval of definition.

- #5

ElijahRockers

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No. That isn't true. Orthogonality depends very much on the particular frequencies and the interval of definition.

We learned in class that {cos(x), cos(2x), ... , cos(Nx)} forms an orthogonal set... is this true?

Not to hijack the thread, but I'm trying to get some visual intuition on how orthogonality depends on the frequency.

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We learned in class that {cos(x), cos(2x), ... , cos(Nx)} forms an orthogonal set... is this true?

Not to hijack the thread, but I'm trying to get some visual intuition on how orthogonality depends on the frequency.

And it depends on the interval. Just because ##\int_{-\pi}^{\pi} \cos(mx)\cos(nx)~dx = 0## if ##m\ne n## doesn't mean, for example, that ##\int_{0}^{\frac \pi 4} \cos(mx)\cos(nx)~dx = 0##. Also, that set of cosines are orthogonal to each other, not to other cosines with different frequencies. So you wouldn't expect, for example, that ##\int_{-\pi}^{\pi} \cos(x)\cos(.9999x)~dx = 0## as you mentioned.

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