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> Fourier Transform became trivial to me when I noticed that it's just a basis transform I always thought this was the definition of Fourier transform, and all

> Fourier Transform became trivial to me when I noticed that it's just a basis transform I always thought this was the definition of Fourier transform, and all the stuff with integrals was just to express the change of basis in an explicit form. What else would the word "transform" mean? How is it even possible for someone to learn about the Fourier transform without it being very explicitly said that it is a change of basis to a different set of orthogonal functions? Did they just introduce the integral expression for it and hope people would memorize it? I don't understand.

j2kun 12y ago
> Did they just introduce the integral expression for it and hope people would memorize it? Unfortunately, yes. The linked article even avoids this language at all costs. And that's just how people work in physics and engineering. They want to talk about spinning functions and all this nonsense instead of just doing some basic linear algebra.
- conistonwater 12y ago
  That's sad. It reminds me of "maths phobia", which is apparently a thing people can have. Also "maths anxiety".
  - j2kun 12y ago
    I think part of it is also that physicists and engineers are the ones primarily teaching the subject. So it might not be that they can't handle the mathematics, but that they simply prefer to think entirely in intuition land where they can spin functions around without anyone protesting.
- SamReidHughe2 12y ago
  That's because it's important that people understand the meaning of the Fourier Transform and what it actually calculates. It's not just some random basis you throw in an equation without understanding what you're doing.
  - conistonwater 12y ago
    You can't "throw a basis in an equation", that's not a known mathematical procedure. Once you learn enough about mathematics, and here linear algebra, you will start to reduce more complicated concepts to more simple concepts to build the complex ideas up from simple ones. In this case, the notion of a change of basis is definitely much simpler than the notion of an integral transform, so it makes sense to express it that way. "What it actually calculates" can be a tricky topic in mathematics. It calculates what we want it to calculate, that's how people came up with the definition. So the way to understand "what it calculates" is to understand "what do I want it to calculate". Somewhat like programming, really. If we want to calculate a function's representation in a convenient basis, then that's what we'll make Fourier transform do.
    - SamReidHughes 12y ago
      Thanks for the response! I've always wanted to have math explained to me by a 21-24 year old that spent too much time with Rudin and Axler in undergrad.
      - cbd1984 12y ago
        Were you planning on bringing anything of substance to this discussion?
- carstimon 12y ago
  Can't we just appreciate that both points of views are helpful? Instead of thinking that one is better, and saying that therefore, the other sucks. Both of them explain different things. The "it's just a basis transformation" makes invertibility, additivity, etc. obvious. The spinning function things makes it clear why we want to do it and what it means to say, approximate by only the small frequencies. (The basis thing tells you nothing about why looking at just the low coefficients is helpful, sometimes. A basis doesn't care about ordering) Except for orientability, I guess.
- carstimon 12y ago
  Can't we just appreciate that both points of views are helpful? Instead of thinking that one is better, and saying that therefore, the other sucks.
- lp251 12y ago
  I agree- the Discrete Fourier Transform can be easily explained in terms of basic linear algebra. And that explanation is usually included in an introduction-level DSP book. Both Oppenheim and Proakis have a section devoted to this view- understanding the DFT as matrix multiplication starts you on the road to the FFT algorithm. But most ECE undergrads will first see the Fourier transform in a continuous setting. Then it isn't so much an issue of "basic linear algebra"- students have almost certainly never talked about the space of square integrable functions, the inner product on this space, a basis for continuous functions, what it means for an infinite collection of objects to be orthogonal, etc... That is, if they have even taken any linear algebra yet! Students learn to associate "Fourier Transform" with "complicated math" and then are discouraged when they learn about the DFT. At least, that has been my experience when TAing signal processing classes.
- nikic 12y ago
  I have no idea about engineering, but I don't see how a physicist would not know this. Basis transformations in infinitely dimensional Hilbert spaces are an integral part of quantum mechanics. The Fourier-transform in particular is used for transformation between space |r> and momentum |p> representations, but it's also common to use energy representations |E_n> where the orthonormal basis is typically somewhat more exotic (e.g. given by a Gaussian multiplied with Hermite polynomials, as is the case for the QHO).
theoh 12y ago
You could learn about it from looking at and understanding a harmonic analyser: http://www.sciencemuseum.org.uk/objects/mathematics/1946-343.aspx http://www.sciencemuseum.org.uk/objects/mathematics/1946-343... Or more likely the reverse mechanism, Kelvin's tide predictor being a good example: http://www.sciencemuseum.org.uk/objects/oceanography/1876-1129.aspx http://www.sciencemuseum.org.uk/objects/oceanography/1876-11... Honestly that is more important to my mental model of the 1D Fourier transform than the algebraic description. It may be slightly inaccurate (are we talking projection or decomposition when calculating the settings on that tide predictor? for example). 2D Fourier transform is more interesting to me as it has applications in graphics. For example, you can synthesise ocean wave-like patterns in a height field by generating a noise image, shaping the noise into a slightly modified disc around the origin, and doing an IFFT. Intuitively you get random wavelength components within a given range, with random orientations. I couldn't write you a formula for that. You can imagine that a 12 year old could understand it, thought. Old paper is here: http://www.computer.org/csdl/mags/cg/1987/03/mcg1987030016.pdf http://www.computer.org/csdl/mags/cg/1987/03/mcg1987030016.p... Another intuitively comprehensible 2D trick is to use an FFT to implement a simulation of fluid dynamics. That's a more complex one but again, it can be understood without algebra. http://www.dgp.toronto.edu/people/stam/reality/Research/pdf/jgt01.pdf http://www.dgp.toronto.edu/people/stam/reality/Research/pdf/...
- darkmighty 12y ago
  I think he meant the 'decomposition' definition. You're decomposing your signal into other signals, that's the fundamental aspect. This decomposition is a linear, so you can also make it so your new components have certain nice properties -- they can be orthogonal, unitary, etc. Then you naturally get waves (complex exponentials) as the components. And then you realize -- hey, those are exponentials -- so the rule that integrating/differentiating each component is trivial allows you to solve linear ODE really quickly. I don't think algebra gets in the way at all -- it just makes this decomposition crystal clear. It makes generalizing easy also -- there are other nice basis like cosine basis, and so on. You can choose any N functions to decompose your N-point function, as long as they are linearly independent, you're guaranteed perfect reconstruction.

j2kun 12y ago

> Did they just introduce the integral expression for it and hope people would memorize it? Unfortunately, yes. The linked article even avoids this language at all costs. And that's just how people work in physics and engineering. They want to talk about spinning functions and all this nonsense instead of just doing some basic linear algebra.

conistonwater 12y ago
That's sad. It reminds me of "maths phobia", which is apparently a thing people can have. Also "maths anxiety".
- j2kun 12y ago
  I think part of it is also that physicists and engineers are the ones primarily teaching the subject. So it might not be that they can't handle the mathematics, but that they simply prefer to think entirely in intuition land where they can spin functions around without anyone protesting.
SamReidHughe2 12y ago
That's because it's important that people understand the meaning of the Fourier Transform and what it actually calculates. It's not just some random basis you throw in an equation without understanding what you're doing.
- conistonwater 12y ago
  You can't "throw a basis in an equation", that's not a known mathematical procedure. Once you learn enough about mathematics, and here linear algebra, you will start to reduce more complicated concepts to more simple concepts to build the complex ideas up from simple ones. In this case, the notion of a change of basis is definitely much simpler than the notion of an integral transform, so it makes sense to express it that way. "What it actually calculates" can be a tricky topic in mathematics. It calculates what we want it to calculate, that's how people came up with the definition. So the way to understand "what it calculates" is to understand "what do I want it to calculate". Somewhat like programming, really. If we want to calculate a function's representation in a convenient basis, then that's what we'll make Fourier transform do.
  - SamReidHughes 12y ago
    Thanks for the response! I've always wanted to have math explained to me by a 21-24 year old that spent too much time with Rudin and Axler in undergrad.
    - cbd1984 12y ago
      Were you planning on bringing anything of substance to this discussion?
carstimon 12y ago
Can't we just appreciate that both points of views are helpful? Instead of thinking that one is better, and saying that therefore, the other sucks. Both of them explain different things. The "it's just a basis transformation" makes invertibility, additivity, etc. obvious. The spinning function things makes it clear why we want to do it and what it means to say, approximate by only the small frequencies. (The basis thing tells you nothing about why looking at just the low coefficients is helpful, sometimes. A basis doesn't care about ordering) Except for orientability, I guess.
carstimon 12y ago
Can't we just appreciate that both points of views are helpful? Instead of thinking that one is better, and saying that therefore, the other sucks.
lp251 12y ago
I agree- the Discrete Fourier Transform can be easily explained in terms of basic linear algebra. And that explanation is usually included in an introduction-level DSP book. Both Oppenheim and Proakis have a section devoted to this view- understanding the DFT as matrix multiplication starts you on the road to the FFT algorithm. But most ECE undergrads will first see the Fourier transform in a continuous setting. Then it isn't so much an issue of "basic linear algebra"- students have almost certainly never talked about the space of square integrable functions, the inner product on this space, a basis for continuous functions, what it means for an infinite collection of objects to be orthogonal, etc... That is, if they have even taken any linear algebra yet! Students learn to associate "Fourier Transform" with "complicated math" and then are discouraged when they learn about the DFT. At least, that has been my experience when TAing signal processing classes.
nikic 12y ago
I have no idea about engineering, but I don't see how a physicist would not know this. Basis transformations in infinitely dimensional Hilbert spaces are an integral part of quantum mechanics. The Fourier-transform in particular is used for transformation between space |r> and momentum |p> representations, but it's also common to use energy representations |E_n> where the orthonormal basis is typically somewhat more exotic (e.g. given by a Gaussian multiplied with Hermite polynomials, as is the case for the QHO).

theoh 12y ago

You could learn about it from looking at and understanding a harmonic analyser: http://www.sciencemuseum.org.uk/objects/mathematics/1946-343.aspx http://www.sciencemuseum.org.uk/objects/mathematics/1946-343... Or more likely the reverse mechanism, Kelvin's tide predictor being a good example: http://www.sciencemuseum.org.uk/objects/oceanography/1876-1129.aspx http://www.sciencemuseum.org.uk/objects/oceanography/1876-11... Honestly that is more important to my mental model of the 1D Fourier transform than the algebraic description. It may be slightly inaccurate (are we talking projection or decomposition when calculating the settings on that tide predictor? for example). 2D Fourier transform is more interesting to me as it has applications in graphics. For example, you can synthesise ocean wave-like patterns in a height field by generating a noise image, shaping the noise into a slightly modified disc around the origin, and doing an IFFT. Intuitively you get random wavelength components within a given range, with random orientations. I couldn't write you a formula for that. You can imagine that a 12 year old could understand it, thought. Old paper is here: http://www.computer.org/csdl/mags/cg/1987/03/mcg1987030016.pdf http://www.computer.org/csdl/mags/cg/1987/03/mcg1987030016.p... Another intuitively comprehensible 2D trick is to use an FFT to implement a simulation of fluid dynamics. That's a more complex one but again, it can be understood without algebra. http://www.dgp.toronto.edu/people/stam/reality/Research/pdf/jgt01.pdf http://www.dgp.toronto.edu/people/stam/reality/Research/pdf/...

darkmighty 12y ago
I think he meant the 'decomposition' definition. You're decomposing your signal into other signals, that's the fundamental aspect. This decomposition is a linear, so you can also make it so your new components have certain nice properties -- they can be orthogonal, unitary, etc. Then you naturally get waves (complex exponentials) as the components. And then you realize -- hey, those are exponentials -- so the rule that integrating/differentiating each component is trivial allows you to solve linear ODE really quickly. I don't think algebra gets in the way at all -- it just makes this decomposition crystal clear. It makes generalizing easy also -- there are other nice basis like cosine basis, and so on. You can choose any N functions to decompose your N-point function, as long as they are linearly independent, you're guaranteed perfect reconstruction.