Timeline for The correct way of looking at Fourier transform
Current License: CC BY-SA 4.0
21 events
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| Feb 29, 2024 at 15:22 | history | edited | Doriano Brogioli | CC BY-SA 4.0 | added 4 characters in body |
| Feb 28, 2024 at 16:01 | comment | added | Abdelrahman | @StevenClark, Oh I've misunderstood you, you're correct in what you are saying, my attempt is to say that Fourier transform correspond to the $$a_{n}$$ in Fourier series $$ f(t) = \sum_n a_n e^{i \omega_0 n t} $$ ,, which describes the frequency content of the original function $f(t)$ | |
| Feb 28, 2024 at 15:37 | comment | added | Steven Clark | @Mans My point was the Fourier series for $f(x)$ as the period goes to infinity is a representation of the function $f(x)$, not the Fourier transform of $f(x)$. This was in response to the last paragraph in the answer above which seemed to claiming the Fourier series for $f(x)$ as the period goes to infinity "becomes similar to a Fourier transform". | |
| Feb 28, 2024 at 14:56 | comment | added | Abdelrahman | @StevenClark, I've never said that the two are equivalent, I'm just saying that in any book I've read considering the Fourier transform, they derived the Fourier transform the same way, they derived the Fourier transform from Fourier series just by taking that the limit as the period goes to infinity, I don't know why you see this as a science fiction, in all the books I've read the derivation is done the same way | |
| Feb 13, 2024 at 20:47 | comment | added | Steven Clark | @Mans With respect to the last paragraph I believe my MSO question on the Fourier inversion theorem illustrates the converse of "this is more science fiction than mathematics". But just to be clear, the Fourier series is not equivalent to the Fourier transform, rather the Fourier series coefficients are derived from the Fourier transform. In the case of a normal Fourier series, the Fourier series coefficients are derived from a truncated Fourier transform (see this answer I recently posted on Math StackExchange). | |
| Jan 6, 2024 at 15:16 | comment | added | Doriano Brogioli | Obtaining the Fourier transform from the Fourier series is only an analogy, a qualitative idea. If you look at the rigorous theorems about Fourier transform, you see that they never rely on the analogy with Fourier series. The reason is exactly what you mention: in Fourier series, you have a discrete sum, while in Fourier transform you have an integral. The exp(i w t) terms are called the basis. There is not much more than this. There is no need to "derive" the Fourier transform: rather, it is useful to prove its properties. | |
| Jan 5, 2024 at 0:50 | comment | added | Abdelrahman | @DorianoBrogioli, I've seen many books and watched lots of videos about the derivation of Fourier transform and it's all about making Fourier transform as the limit of Fourier series, and I Know that the Fourier series is written as the summation of infinite basis,but what does it mean that Fourier transform is integral of the basis ? , isn't the basis must be n-dimensional ? | |
| Jan 3, 2024 at 15:30 | comment | added | Doriano Brogioli | It was nice to discuss with you. I hope that my answer was useful. | |
| Jan 3, 2024 at 14:37 | comment | added | Doriano Brogioli | Yes, this is what I think. Of course, it is more an opinion than a rigorous concept! Typically, the topic is explained in books on analysis, in particular where the Hilbert spaces are discussed. However, there are specific books like this: "Fourier and Laplace Transforms", Beerends, Morsche, van den Berg, van de Vrie. | |
| Jan 3, 2024 at 14:34 | comment | added | Abdelrahman | @DorianoBrogioli,so the main idea of Fourier transform is representing function by some basis, looking at Fourier transform as limit of Fourier series is just another way of looking at it, but the main way of thinking about Fourier transform is representing function in some basis by projecting it, is that correct? | |
| Jan 3, 2024 at 14:31 | comment | added | Abdelrahman | @DorianoBrogioli, so all the transform (Laplace or Fourier) are just representing function by some basis, can you provide me a book that talks in detail about this subject and help me to understand how this transformation works | |
| Jan 3, 2024 at 14:29 | comment | added | Doriano Brogioli | Now, the question can be: "what is the relation between Fourier series and anti-transform?" Then, the answer is that, intuitively, and roughly, you reach the Fourier anti-transform as the limit of the Fourier series. But I feel that the two questions are separate and different. | |
| Jan 3, 2024 at 14:28 | comment | added | Doriano Brogioli | For sure, the Fourier anti-transform, the Laplace anti-transform and the Fourier series represent some functions (not all the functions) as sums (or integrals) of some basis. The Fourier has the advantage, over the Laplace transform, that the basis is orthonormal, so the transform and the anti-transform are well defined and very similar. | |
| Jan 3, 2024 at 14:26 | comment | added | Abdelrahman | @DorianoBrogioli, so the main idea of Laplace and Fourier transform is just projection of function on some basis , not that Fourier transform came from Fourier series | |
| Jan 3, 2024 at 14:23 | comment | added | Doriano Brogioli | The Laplace transform is somehow similar to the Fourier transform. It also represents the function as a sum (or integral) of some basis. However, in the case of Laplace transform, the basis is not orthogonal. | |
| Jan 3, 2024 at 14:22 | comment | added | Abdelrahman | @LL3.14, I really have a confusion, I know that Fourier transform represent our function by just a bunch of sines and I Know how he did that by considering transform generalization of the Fourier series, but I didn't know how does Laplace represent a function as exponential and sinusoids, so I searched and found something called projection of a function of a basis and I think about like a projection of a vector, so I rethought about Fourier transform and said it just a projection and so like Laplace transform, but this doesn't seem right, so can you provide me a derivation for Laplace transfor | |
| Jan 3, 2024 at 13:48 | comment | added | LL 3.14 | In the Laplace transform, there is no orthonormal basis .. | |
| Jan 3, 2024 at 13:35 | comment | added | Abdelrahman | So you say the correct way of looking at Fourier transform is just by projecting our main function on the basis $e^{-iwt}$ and by doing that we see the frequency inside our original function? And this apply to any transform like Laplace transform because there is no Laplace series to derive from it Laplace transform | |
| Jan 3, 2024 at 13:08 | comment | added | Doriano Brogioli | Well, maybe my comment on "science fiction" is a bit too much, but it is true that there are lots of subtleties more than bringing the period to infinite. For example: what is the space of functions that you can describe? For example, not $x^2$. So, I agree that you can intuitively go from the series to the integral by increasing the period, but I think that this fact does not help us too much. | |
| Jan 3, 2024 at 13:02 | comment | added | Abdelrahman | But the derivation I did know about Fourier transform is just by deriving it from the Fourier series by taking the period to tend to infinity | |
| Jan 3, 2024 at 12:54 | history | answered | Doriano Brogioli | CC BY-SA 4.0 |