Talk:Fourier series

Error in Hilbert space section?

It seems to me that the Hilbert space being considered must be $L^{2}(-\pi ,\pi )$, since otherwise we need $f(-\pi )=f(\pi )$ by periodicity. jajaperson (talk) 05:25, 6 March 2024 (UTC)

"Hilbert Spaces and Fourier analysis" listed at Redirects for discussion

The redirect Hilbert Spaces and Fourier analysis has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2024 March 28 § Hilbert Spaces and Fourier analysis until a consensus is reached. 1234qwer1234qwer4 01:32, 28 March 2024 (UTC)

Continuously differentiable functions

Back in 2008 in this edit User:Silly rabbit put "We have already mentioned that if ƒ is continuously differentiable, then $n{\hat {f}}(n)$ converges to zero as n goes to infinity. It follows, essentially from the Cauchy-Schwarz inequality, that the Fourier series of ƒ converges absolutely. Since the Fourier series converges in the mean to ƒ, it must converge uniformly:" and then he states a theorem.

I don't understand what is meant. (I tried sending an e-mail but got no answer.) What does this have to do with the Cauchy-Schwarz inequality? How do you use that to show this?

It is certainly possible for a sequence of coefficients to go to zero, and for n times them to go to zero, without the sequence being summable. For example, if $a_{n}=1/(n\log n)$ (for $n>1$) then they go toward zero, as does $na_{n}$, but the sum of the $a_{n}$'s goes to infinity.

Also, what does "converges in the mean" mean?

User:Charles Matthews, maybe you can help?

Eric Kvaalen (talk) 16:10, 7 September 2024 (UTC)

I did reply to the email. The key point is that if $f$ is continuously differentiable, the sequence Fourier coefficients of $f'$, $n{\hat {f}}(n)$, belongs to $\ell ^{2}$. Therefore, ${\hat {f}}(n)$ belongs to $\ell ^{1}$, because (by Cauchy)

$\left(\sum |{\hat {f}}(n)|\right)^{2}\leq \sum {\frac {1}{n^{2}}}\cdot \sum |n{\hat {f}}(n)|^{2}.$

Converges in mean is convergence in $L^{1}$. -- Silly rabbit (talk) 16:47, 7 September 2024 (UTC)

@Silly rabbit: Ah, I found your e-mail. Stupid Hotmail put it in my Junk folder! Thanks for the explanation.

So here's another question: I see no theorem that says that the Fourier series for a function that is zero at zero and $x\sin(1/x)$ elsewhere between $-\pi$ and $\pi$ converges everywhere. Do we know whether it converges at zero? Eric Kvaalen (talk) 20:19, 8 September 2024 (UTC)

I don't know. $x\sin(1/x)$ is 1/2-Holder continuous, which is not quite enough for uniform convergence. For instance $x^{1+\epsilon }\sin(1/x)$ has a uniformly convergent Fourier series for $\epsilon >0$. Silly rabbit (talk) 23:28, 8 September 2024 (UTC)

@Silly rabbit: OK. Anyway, $x^{1+\epsilon }\sin(1/x)$ is of bounded variation in the interval, so the series converges by the Dirichlet–Jordan test. It seems to me that it would be difficult to show that $x\sin(1/x)?$ does not converge at zero. What about the derivative of $x\sin(1/x)?$ Its Fourier coefficients are well defined, but does it converge anywhere? Eric Kvaalen (talk) 10:13, 9 September 2024 (UTC)

Actually, the example of the Fourier series of $x\sin(1/x)$ does converge to 0 at $x=0$, because the function does satisfy a Holder condition (of exponent $0<\alpha <1/2$), by the Dini test. Silly rabbit (talk) 11:29, 9 September 2024 (UTC)

@Silly rabbit: Aha, you're right. (I think we can say $\alpha =1/2.$) However, that doesn't apply to its derivative. Eric Kvaalen (talk) 08:07, 10 September 2024 (UTC)

@Silly rabbit: Are there any continuously differentiable functions that are not already covered by being of bounded variation, so the Dirichlet-Jordan theorem applies? Eric Kvaalen (talk) 09:15, 16 September 2024 (UTC)

The Table of common Fourier series would be much clearer

The Table of common Fourier series would be much clearer if *instead* of labeling the columns with "time domain" and "frequency domain", they were labeled with "Function" and "Fourier coefficients".

For several reasons:

1. Above all, that is overwhelmingly how most people think: in terms of functions and their Fourier coefficients. So why be confusing?

2. It is entirely unclear why the column of functions (of whose Fourier coefficients we will soon be told) is labeled with a *domain*, since it is a function, not a domain.

3. There are many, many applications of Fourier series, since they apply to almost any real-valued function that arises in practice. A very large number of those applications are not about either time or frequency.

I have no objection, of course, to mentioning the reasons that the time domain and frequency domain are highly relevant to Fourier analysis, even in this table ... but my point is: Make it easy for people to find words like "function" and "Fourier coefficients", since that is mainly what they will be looking for.

— Preceding unsigned comment added by 98.36.148.11 (talk • contribs)

While I don't love the "time/frequency" metaphor, it is baked into the structure of the article and really does clarify a lot of things here. So I'm opposed to any plan to eliminate it from one part of the article, without a provision for improving the rest of the article without sacrificing clarity of the presentation. Tito Omburo (talk) 10:16, 8 November 2024 (UTC)

Symmetry relations

@Tito Omburo, I see you've reverted my edit regarding the deletion of the duplicate "Symmetry" subsection.

The only difference is the symbol $s$ instead of $f$, the subsections Fourier_series#Symmetry_relations and Fourier_transform#Symmetry are otherwise indistinguishable. I think it's rather presumptuous to state that it's the Fourier transform beteeen Z and S^1.

Could you please elaborate on why you feel the subsections are distinct?

Kind regards, Roffaduft (talk) 12:11, 12 December 2024 (UTC)

The Fourier transform covered in the article Fourier transform is for continuous signals, while the symmetry relations covered here are applied for discrete signals or periodic signals. For example, the Fourier transform is gotten by integrating over R, while Fourier series are gotten by integrating over S^1. (Although often in digital signal processing, the roles of time and frequency are reversed, and the signal is discrete, and the Fourier transform is periodic.) Tito Omburo (talk) 12:39, 12 December 2024 (UTC)

Yes, I understand the difference, but nowhere in the subsection is this distinction being made. It is very cleary a simple copy/paste with no information specific to the Fourier series.

If the symmetry subsection was written as in

• Oppenheim and Schafer (1975) Digital Signal Processing, Prentice–Hall, pp 20–26.

then I wouldn't have deleted it. Roffaduft (talk) 12:46, 12 December 2024 (UTC)

Recent edits

The 11 Dec 2024 version was careful to distinguish between real-valued s(x) and complex-valued s(x). The analysis and synthesis formulas are developed for real-valued s(x). Eq.6, for instance, states

$c_{_{n}}=c_{|n|}^{*},\quad n<0$

which is true in general only for real-valued s(x). There was once a section to explain the generalization to complex s(x). Here is a copy/paste version:

Complex-valued functions

If $s(x)$ is a complex-valued function of a real variable $x,$ both components (real and imaginary part) are real-valued functions that can be represented by a Fourier series. The two sets of coefficients and the partial sum are given by:

$c_{_{Rn}}={\frac {1}{P}}\int _{P}\operatorname {Re} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx$ and $c_{_{In}}={\frac {1}{P}}\int _{P}\operatorname {Im} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx$

$s_{_{N}}(x)=\sum _{n=-N}^{N}c_{_{Rn}}\cdot e^{i{\tfrac {2\pi }{P}}nx}+i\cdot \sum _{n=-N}^{N}c_{_{In}}\cdot e^{i{\tfrac {2\pi }{P}}nx}=\sum _{n=-N}^{N}\left(c_{_{Rn}}+i\cdot c_{_{In}}\right)\cdot e^{i{\tfrac {2\pi }{P}}nx}.$

Defining $c_{n}\triangleq c_{_{Rn}}+i\cdot c_{_{In}}$ yields:

$s_{_{N}}(x)=\sum _{n=-N}^{N}c_{n}\cdot e^{i{\tfrac {2\pi }{P}}nx}$ (Eq.7)

This is identical to Eq.5 except $c_{n}$ and $c_{-n}$ are no longer complex conjugates. The formula for $c_{n}$ is also unchanged:

${\begin{aligned}c_{n}&={\tfrac {1}{P}}\int _{P}\operatorname {Re} \{s(x)\}\cdot e^{-i{\frac {2\pi }{P}}nx}\ dx+i\cdot {\tfrac {1}{P}}\int _{P}\operatorname {Im} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx\\[4pt]&={\tfrac {1}{P}}\int _{P}\left(\operatorname {Re} \{s(x)\}+i\cdot \operatorname {Im} \{s(x)\}\right)\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx\ =\ {\tfrac {1}{P}}\int _{P}s(x)\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx.\end{aligned}}$

end

I added that section, but later replaced it with a short statement and footnote, which are now gone. Bob K (talk) 13:30, 13 December 2024 (UTC)

Not sure where to start on this one. First of all, in the definition of the Fourier series (at the beginning of the subsection), $s$ and $c$ are complex. I've added an inline citation to Folland (1992) to support this statement.

Second, the references you've provided do not make any distinction between real- and complex-valued $s(x)$, for good reason. Because there really isn't any.

Third, in the complex case, $c_{n}={\tfrac {1}{2}}(a_{n}-ib_{n})$ and $c_{-n}={\tfrac {1}{2}}(a_{n}+ib_{n})$. The only thing that changes is that $a_{n}$ and/or $b_{n}$ become complex.

Fourth, there was no clear distinction between the real and complex subsection. That is, the statements in the "real" subsection applied just as well to the complex case.

Fifth, the entire "complex valued functions" section you posted here is basically a very convoluted way of showing there is no difference between the real and complex case, i.e., Eq.7 = Eq.5.

Kind regards, Roffaduft (talk) 14:31, 13 December 2024 (UTC)

Strictly speaking, I do not think it's true that every complex signal has an amplitude-phase form (using a real sinusoid ad the basic wavelet). In general, the real and imaginary parts come in different phases. With an amplitude-phase form, one always has I think <math<|c_{-n}|=|c_n|</math>. So it's not exactly true that there is no difference between the two cases. (But I dont disagree that, broadly, the specific passage seems unnecessary.) Tito Omburo (talk) 16:20, 13 December 2024 (UTC)