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the way Fourier transform$$f(ξ) = \int_{-∞}^{∞} f(x) e^{-2πixξ}dx$$ describe the frequency extent of the function is derived from the Fourier series by taking that $c_{n}$ of the Fourier series and make the period goes to infinity, so the ${c_n}$ is equivalent to $f(ξ)$ but $c_{n}$ for discrete frequencies and $f(ξ)$ for continuous of frequencies, that what I Know and that what all the books that I've read from derive the Fourier transform,

but when I discuss with many users about this connection between Fourier series and Fourier transform , they just say this connection is not totally rigorous and the way Fourier transform is defined isn't by Limiting the Fourier series , I don't know what their point really is but I don't what's wrong with what I am saying all the books that I read from derive it the same way by the same procedure, if so what is the objection, then?

I've read in a paper concerning this part that by integrating our $f(x)$ with this periodic function $e^{-2πixξ}$ it gives us the information about the frequency extent of our $f(x)$, but I don't understand what is their reason, they just say this.

  • The idea of Fourier series, describing periodic functions on intervals of a certain length $L$, can indeed be extended to the real axis. But other (more modern) approaches to the Fourier transform may also be valid. Regarding your last question, $f(x)$ can be considered to be the sum of many oscillatory terms. By multiplying with the phase factor and integrating over$x$ one obtains the amplitude of one single frequency. That is what is meant. The Fourier transform acts as a narrow filter, like a radio which is tuned to one frequency. – M. Wind Mar 26 '24 at 15:21
  • About the filtering property: I suggest you take a look at the Dirac delta function. – M. Wind Mar 26 '24 at 15:29
  • @M.Wind, Can you give any book that talks about the other approaches to Fourier transform as you mentioned – Abdelrahman Mar 26 '24 at 15:38
  • @M.Wind, Can you explain to me how multiplying by the phase factor and integrating over $x$ make us obtain the $f( ξ)$ ? because I didn't get it – Abdelrahman Mar 26 '24 at 15:40
  • I don't really have the time to write a fleshed out answer at the moment (and, since it has been a while since I've played with the theory much, I am not even confident that I could do a good job), but the underlying idea, in a view-from-1000-miles-up kind of way, is Pontryagin duality. Note that for both the transform and series, characters are mapped to elements of the Pontryagin dual of some space, and summing (or integrating) over all of the characters gives the transform/series. – Xander Henderson Mar 26 '24 at 18:33
  • Katznelson has a moderately approachable text on the more general theory of harmonic analysis. – Xander Henderson Mar 26 '24 at 18:33
  • Perhaps this YouTube video answers your question, but it seems to me this is really derivation of the inverse Fourier transform $$fx)=\frac{1}{2 \pi} \int\limits_{-\infty}^\infty \hat{f}(\omega), e^{i x \omega} , d\omega$$ from the Fourier series where the Fourier transform $\hat{f}(\omega)$ of $f(x)$ is defined as $$\hat{f}(\omega)=\int\limits_{-\infty}^\infty f(x), e^{-i \omega x} , dx.$$ – Steven Clark Jun 06 '24 at 23:02

1 Answers1

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This question seems related to your other questions The correct way of looking at Fourier transform and How the Fourier transform show the frequency extent of $f(x)$?.


A function $f(x)$ can be approximated on the interval $-\frac{P}{2}<x<\frac{P}{2}$ by the exponential Fourier series

$$f(x)=\sum\limits_{n=-\infty}^\infty c_n\, e^{i 2 \pi \frac{n}{P} x}\tag{1}$$

where

$$c_n=\frac{1}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)\, e^{-i 2 \pi \frac{n}{P} x} \, dx\tag{2}$$


The Fourier transform of the function $f(x)$ can be defined as

$$F(\omega)=\mathcal{F}_{x}[f(x)](\omega)=\lim\limits_{P\to\infty}\left(\int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)\, e^{-i 2 \pi \omega x} \, dx\right)\tag{3}$$

which is somewhat analogous to formula (2) above (the exact relationship is clarified further below).


Also the Fourier series for $f(x)$ and the inverse Fourier transform

$$f(x)=\mathcal{F}^{-1}_{\omega}[F(\omega)](x)=\int\limits_{-\infty}^{\infty} F(\omega)\, e^{2 \pi i x \omega} \, d\omega\tag{4}$$

both recover the function $f(x)$, so the Fourier series for $f(x)$ is more analogous to the the inverse Fourier transform than the Fourier transform.


I think the correct way of looking at the connection between the inverse Fourier transform and Fourier series is the "nested" Fourier series representation

$$f(x)=\mathcal{F}^{-1}_{\omega}[F(\omega)](x)=\int\limits_{-\infty}^{\infty} F(\omega)\, e^{2 \pi i x \omega} \, d\omega\\=\lim\limits_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{1}{2} \sum\limits_{k=-2 f (2 n-1)}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right) e^{\frac{i \pi k x}{2 n-1}}\\-\frac{1}{8} \sum\limits_{k=-4 f (2 n-1)}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right) e^{\frac{i \pi k x}{4 n-2}}\right)\right)\tag{5}$$

where $\mu(n)$ is the Möbius function, the evaluation frequency $f$ in the inner sum over $k$ is assumed to be a positive integer, and

$$F(\omega)=\mathcal{F}_{x}[f(x)](\omega)=\int\limits_{-\infty}^{\infty} f(x)\, e^{-i 2 \pi \omega x} \, dx\tag{6}$$

is the Fourier transform of $f(x)$.


Formula (5) above is a refinement of formula (5) in my related MSO question which provides information on its derivation.


I believe the "nested" Fourier series for $f(x)$ defined in formula (5) above converges for $x\in\mathbb{R}$ when the function $f(x)$ is recoverable from its Fourier transform $F(\omega)$ via the Fourier inversion theorem. My related MSO question linked above illustrates several examples of recovering a function $f(x)$ from its Fourier transform $F(\omega)$ via the "nested" Fourier series representation defined in formula (5) above.


The remainder of this answer clarifies more precisely the relationship between the Fourier series coefficients $c_n$ defined in formula (2) above and the Fourier transform $F(\omega)$ defined in formula (6) above, and also similarities and differences between the Fourier series and the "nested" Fourier series defined in formulas (1) and (5) above.


Note the Fourier series coefficient $c_n$ in formulas (1) and (2) above can be evaluated as

$$c_n=\frac{1}{P}\, C_P\left(\frac{n}{P}\right)\tag{7}$$

where

$$C_P(\omega)=\int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x) \, e^{-i 2 \pi \omega x} \, dx\tag{8}$$

is a Fourier transform with a truncated integration range and note that

$$F(\omega)=\lim\limits_{P->\infty} C_P(\omega)\tag{9}$$


Note the functions $F(\omega)$ and $C_P(\omega)$ are both continuous, but since $F(\omega)$ is only evaluated at rational values of $\omega$ in formula (5) above its only necessary to account for a countably infinite number of frequencies to recover the function $f(x)$ from the "nested" Fourier series in formula (5) above as well as the Fourier series in formula (1) above.


It seems to me that saying $F(\omega)$ represents the frequency content of $f(x)$ is analogous to saying that $C_P(\omega)$ represents the frequency content of $f(x)$, but note there are an uncountably infinite number of values of $\omega$ for which $C_P(\omega)\ne 0$ which are not accounted for in the Fourier series for $f(x)$, and likewise there are an uncountably infinite number of values of $\omega$ for which $F(\omega)\ne 0$ which are not accounted for in the "nested" Fourier series for $f(x)$. So perhaps its more correct to say $F(\omega)$ contains the frequency content of $f(x)$ in the same sense that $C_P(\omega)$ contains the frequency content of $f(x)$.


Fourier series can be used to approximate periodic functions, but the "nested" Fourier series representation is not applicable to periodic functions since the the Fourier transform of a periodic function doesn't converge in the usual sense. Fourier series can be used to approximate a function over some interval of $x$, whereas the "nested" Fourier series can be used to recover a function for $x\in\mathbb{R}$ when the Fourier inversion theorem is applicable (and even for $x\in\mathbb{C}$ when $f(x)$ is holomorphic).

Steven Clark
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  • In your answer you said that Fourier transform is analogous to $c_{n}$ and inverse Fourier transform is analogous to Fourier series, Are you saying that it's just another way for looking at Fourier transform and inverse Fourier transform and not the main way or the original derivation? – Abdelrahman Mar 26 '24 at 18:21
  • If you have any good book that talks about the material from the beginning to solve my confusion, I would be greatful – Abdelrahman Mar 26 '24 at 18:26
  • @Mans I don't believe you'll find any books that talk the relationship in formula (5) above since I recently derived this result, but I believe formula (5) above provides a new insight into the relationship between the inverse Fourier transform and Fourier series which perhaps can help clear up some of your confusion. I believe many users posting answers to questions like this on Math StackExchange and many authors of books like the ones you're reading are also confused on this topic which is perhaps why you're not able to make sense of what they're saying. – Steven Clark Mar 26 '24 at 18:39
  • @Mans I'm saying when a function is recoverable from its Fourier transform via the Fourier inversion theorem, the inverse Fourier transform defined in formula (6) in my answer above can be evaluated as the nested Fourier series representation defined in formula (5) in my answer above. I haven't really looked at the derivation you're referring to since you didn't provide it in your question. – Steven Clark Mar 26 '24 at 18:47
  • I'll try to provide to you the derivation I know for the Fourier transform to show you my point – Abdelrahman Mar 26 '24 at 18:56