[math-fun] Periodic functions & Fourier transforms ?
I learned about periodic functions & Fourier transforms while an undergraduate. In particular, I learned that the Fourier transform of a periodic function was a countable set of complex numbers representing the Fourier coefficients of evenly spaced impulse "functions". These coefficients represented the amplitude and phase of each of the "overtones" of the fundamental frequency of the periodic function. The only problem with this analysis is that we have taken an *uncountable* set of real numbers -- i.e., the periodic function values over one period -- and represented them by a *countable* set of complex numbers -- the Fourier coefficients. I would guess that the problem is that nearly all such periodic functions *can't* be represented by such a Fourier transform, perhaps because "almost all" such functions have too many *discontinuities*; i.e., they aren't sufficiently "smooth". If there were a countable number of discontinuities, that function could still conceivably be represented by a Fourier series. But once the number of discontinuities becomes uncountable, we're out of luck? Would this explanation be a reasonable intuition? (No wonder Cantor went mad!)
"Fourier series" is what you are looking for. The "number of" periodic functions is of course greatly reduced by considering only, say, piece-wise continuous(ly differentiable) functions. Then there are (IIRC) two notions of convergence important here, point-wise (PC) and uniform (UC). You get UC for continuous functions..., oh, wait: https://en.wikipedia.org/wiki/Convergence_of_Fourier_series Best regards, jj * Henry Baker <hbaker1@pipeline.com> [Dec 21. 2015 07:44]:
I learned about periodic functions & Fourier transforms while an undergraduate.
In particular, I learned that the Fourier transform of a periodic function was a countable set of complex numbers representing the Fourier coefficients of evenly spaced impulse "functions". These coefficients represented the amplitude and phase of each of the "overtones" of the fundamental frequency of the periodic function.
The only problem with this analysis is that we have taken an *uncountable* set of real numbers -- i.e., the periodic function values over one period -- and represented them by a *countable* set of complex numbers -- the Fourier coefficients.
I would guess that the problem is that nearly all such periodic functions *can't* be represented by such a Fourier transform, perhaps because "almost all" such functions have too many *discontinuities*; i.e., they aren't sufficiently "smooth".
If there were a countable number of discontinuities, that function could still conceivably be represented by a Fourier series. But once the number of discontinuities becomes uncountable, we're out of luck?
Would this explanation be a reasonable intuition?
(No wonder Cantor went mad!)
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On 2015-12-20 23:06, Joerg Arndt wrote:
"Fourier series" is what you are looking for. The "number of" periodic functions is of course greatly reduced by considering only, say, piece-wise continuous(ly differentiable) functions.
Then there are (IIRC) two notions of convergence important here, point-wise (PC) and uniform (UC). You get UC for continuous functions..., oh, wait: https://en.wikipedia.org/wiki/Convergence_of_Fourier_series
Best regards, jj
I'm too lazy to check whether this article blesses http://www.tweedledum.com/rwg/cog.htm (Click to slightly enlarge.) --rwg
* Henry Baker <hbaker1@pipeline.com> [Dec 21. 2015 07:44]:
I learned about periodic functions & Fourier transforms while an undergraduate.
In particular, I learned that the Fourier transform of a periodic function was a countable set of complex numbers representing the Fourier coefficients of evenly spaced impulse "functions". These coefficients represented the amplitude and phase of each of the "overtones" of the fundamental frequency of the periodic function.
The only problem with this analysis is that we have taken an *uncountable* set of real numbers -- i.e., the periodic function values over one period -- and represented them by a *countable* set of complex numbers -- the Fourier coefficients.
I would guess that the problem is that nearly all such periodic functions *can't* be represented by such a Fourier transform, perhaps because "almost all" such functions have too many *discontinuities*; i.e., they aren't sufficiently "smooth".
If there were a countable number of discontinuities, that function could still conceivably be represented by a Fourier series. But once the number of discontinuities becomes uncountable, we're out of luck?
Would this explanation be a reasonable intuition?
(No wonder Cantor went mad!)
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Henry Baker -
Joerg Arndt -
rwg