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It seems that some, especially in electrical engineering and musical signal processing, describe that every signal can be represented as a Fourier series.

So this got me thinking about the mathematical proof for such argument.

But even after going through some resources about the Fourier series (which I don't have too much background in, but grasp the concept), I cannot find a mathematical proof for whether every function can be represented by a Fourier series. There was a hint about the function having to be periodic.

So that means that the "every function can be represented as a Fourier series" is a myth and it doesn't apply on signals either, unless they're periodic?

But then I can also find references like these: http://msp.ucsd.edu/techniques/v0.11/book-html/node171.html that say/imply that every signal can be made periodic? So does that change the notion about whether Fourier series can represent every function, with the new condition of first making it periodic, if necessary?

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  • $\begingroup$ Yes, a function that has a Fourier series must be periodic. There are further conditions. $\endgroup$ – Zhen Lin Jul 30 '15 at 0:34
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    $\begingroup$ A straightforward counting argument shows that most functions cannot be represented as a Fourier series. A Fourier series is determined by a countable family of Fourier coefficients, so the set of such series has cardinality ${\mathfrak c}^{\aleph_0}$, whereas the set of real valued functions defined on some interval has cardinality ${\mathfrak c}^{\mathfrak c}$. $\endgroup$ – MJD Jul 30 '15 at 0:44
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    $\begingroup$ Every function given on an interval $[a,b]$ can be made periodic outside. Periodicity is not a trouble for signals, they have a finite life time, so nobody cares what's outside. $\endgroup$ – A.Γ. Jul 30 '15 at 0:44
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    $\begingroup$ Every function in $L^2([a,b])$ can be uniquely represented in the sense of $L^2$ by a Fourier series. The deeper fact is Carleson's theorem, which was one of the most difficult achievements in 20th century analysis, and tells us about the precise conditions for pointwise (actually, "pointwise almost everywhere") convergence of Fourier series. $\endgroup$ – Ian Jul 30 '15 at 0:53
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    $\begingroup$ If you're going to sample the signal over an interval $T$ of time, then the Fourier series with exponentials $e^{i2n\pi/T}$ for $n=0,\pm 1,\pm 2,\cdots$ will represent the signal over $[0,T]$. Obviously outside of that interval $[0,T]$, the Fourier series extends periodically, even though the signal may do something else. $\endgroup$ – Disintegrating By Parts Jul 30 '15 at 20:35
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Since you're referring to signals here, it seems appropriate to consider this question from the viewpoint of an electrical engineer.

If we impose some restrictions on what kind of functions can be considered a "signal," then all periodic signals have a Fourier series.

  • The function should be piecewise continuous.
  • the function should be be bounded.

These are reasonable physical restrictions that all real signals should meet. These are also more than enough for a function to have a Fourier series.

Now, for a function that isn't periodic, we can find a Fourier series for a piece of it through a process called "windowing." Basically you isolate a part of the signal on some interval, and pretend that piece is one period of a periodic signal. The Fourier coefficients for each "window" tell you the power spectrum of the signal as time progresses.

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    $\begingroup$ Does pretending something to be periodic lead to a "lossy" Fourier series? That is, the original signal cannot be reconstructed by the inverse FFT? $\endgroup$ – mavavilj Jul 30 '15 at 9:48
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I came across this question because I wanted to ask the same thing. In Gilbert Strang's Linear Algebra LEC 24, towards the end: https://youtu.be/8MF3pz-oYHo?t=41m8s he mentions that the parts of a fourier series is like an orthogonal basis, and that you can project a function onto the fourier series like any other projection onto orthogonal basis.

And the intuitive way to think about its restriction to be periodic is because the parts that make up the fourier series are periodic.

(I'd love to see something more rigourous though)

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A bounded periodic integrable function F will certainly "have" a Fourier series, but the sum of the series can fail to be equal to F at some points, even if F is continuous.

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Well there are 3 conditions for a Fourier Series of a function to be exist: 1. It has to be periodic. 2. It must be single valued, continuous.it can have finite number of finite discontinuities. 3. It must have only a finite number of Maxima and minima within the period. 4. The Integral over one period of |f(X)| must converge. Each of them have Analytical proofs but let's discuss them using analogy. 1). See that Fourier series can be written as complex exponent and they represent circles(like this one https://www.youtube.com/watch?v=55y13PF0uSg&feature=share ), see as circles are bounded curve , the vectors connected with them and the output point must come back on the same path for a larger time , hence they must be periodic (well for non-periodic functions it also works but you will only get the result for a particular region). 2).The continuous part is clear I hope so. For the discontinuous part, you have to remember that they never go to infinity as if it does than at that part the curve goes forever and as circular motions are periodic , the expansion must have a finite bound (in a region) In a similar way we can get the Intitution of other 2 points.

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  • $\begingroup$ This answer is hard to follow, but it's a lot closer to what I was looking for when I searched for this question than the other answers are. I watched the 3Blue1Brown video where he said, "Any function can be expressed as a sum of sign waves. Well, a pedantic mathematician might point out that there are some pathological exceptions — certain weird functions where this isn't true, but basically...", and I wanted to know what those "pathological exceptions" were. $\endgroup$ – H. H. Jun 26 '20 at 9:15
  • $\begingroup$ Weierstrass functions are expressed as a Fourier series, and they have an infinite number of maxima and minima between any two points, so I guess your 3 conditions are sufficient to prove a function has a Fourier series but not necessary for it to have one? $\endgroup$ – H. H. Jun 26 '20 at 9:23

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