3 ms·

I see this particular mistake crop up a lot, and it's a bit pedantic, but I don't get the impression that it's being intentionally ignored for educational reaso

by hansvm 5y ago

I see this particular mistake crop up a lot, and it's a bit pedantic, but I don't get the impression that it's being intentionally ignored for educational reasons: The author repeatedly describes "the" recipe, "the" set of frequencies, .... The standard Fourier transform is only unique conditioned on the fact that we've agreed on a particular set of basis functions ahead of time. It isn't even the only choice of orthogonal exponentials we could have picked. In terms of real applications that point often doesn't matter much -- especially in fields like lossy compression where you might not actually care much about any underlying meaning in the representation, and especially in fields like radio where your data is in some sense fundamentally represented as combinations of sinusoids -- but in general it isn't a sound leap to go from "this Fourier coefficient is big" to "there exists a big sinusoid that semantically matters to my data."

whatshisface 5y ago
Linear time-invariant physical processes are so common in the real world that it's not unreasonable to expect sinusoids to semantically matter... At least when the data pertains to physical systems.
- hansvm 5y ago
  Certainly. It's super reasonable to think that they might matter and to investigate that approach; it's unreasonable to take as an axiom that they matter and use that as proof of other facts.
joppy 5y ago
If you add some additional structure of remembering that the domain of these transformations is an abelian group, then there is actually a completely canonical version of the Fourier transform, whose domain is the dual group - the only non-canonical part is picking labels for elements of the dual group, but there is no other choice of basis functions. (Wikipedia gets a little into this, you can find it under "Pontryagin duality" on the Fourier transform page, but it's not really enough of an explanation to learn from). Take the discrete Fourier transform for example, whose domain group is Z/nZ: the integers-mod-n for some fixed n. The basis functions for the transforms need to be group homomorphisms f: Z/nZ -> C, meaning that they need to satisfy f(a + b) = f(a) f(b) for all a, b. Without too much trouble you can work out that the only such functions are of the form f(k) = exp(2 pi i m k / n) for m = 0, ..., n - 1. So once the group structure is remembered, the choice of basis is forced. The only ambiguity is relabelling the basis elements back into Z/nZ, which is a (very useful) hack. The same reasoning (with more complicated working) applies to taking the Fourier series of a periodic function, or taking the Fourier transform of a continuous function on the real line. There is a canonical choice of basis, but some ambiguity in what labels to assign to those basis elements.
- whatshisface 5y ago
  Since you could be in any basis before taking the Fourier transform, it's more like there's a canonical change-of-basis than there is a canonical basis. Time domain might not be the system's most natural basis.

by hansvm 5y ago

whatshisface 5y ago

Linear time-invariant physical processes are so common in the real world that it's not unreasonable to expect sinusoids to semantically matter... At least when the data pertains to physical systems.

hansvm 5y ago
Certainly. It's super reasonable to think that they might matter and to investigate that approach; it's unreasonable to take as an axiom that they matter and use that as proof of other facts.

joppy 5y ago

If you add some additional structure of remembering that the domain of these transformations is an abelian group, then there is actually a completely canonical version of the Fourier transform, whose domain is the dual group - the only non-canonical part is picking labels for elements of the dual group, but there is no other choice of basis functions. (Wikipedia gets a little into this, you can find it under "Pontryagin duality" on the Fourier transform page, but it's not really enough of an explanation to learn from). Take the discrete Fourier transform for example, whose domain group is Z/nZ: the integers-mod-n for some fixed n. The basis functions for the transforms need to be group homomorphisms f: Z/nZ -> C, meaning that they need to satisfy f(a + b) = f(a) f(b) for all a, b. Without too much trouble you can work out that the only such functions are of the form f(k) = exp(2 pi i m k / n) for m = 0, ..., n - 1. So once the group structure is remembered, the choice of basis is forced. The only ambiguity is relabelling the basis elements back into Z/nZ, which is a (very useful) hack. The same reasoning (with more complicated working) applies to taking the Fourier series of a periodic function, or taking the Fourier transform of a continuous function on the real line. There is a canonical choice of basis, but some ambiguity in what labels to assign to those basis elements.

whatshisface 5y ago
Since you could be in any basis before taking the Fourier transform, it's more like there's a canonical change-of-basis than there is a canonical basis. Time domain might not be the system's most natural basis.