Talk:Fourier series

Contents

1 Inconsistencies 2 Cis function and rewrite 3 The fourier series of an arbitrary periodic continuous function need not converge to the function. 4 Constructiveness 5 Generalized to interval a-b 6 How to compose good-looking non-TeX mathematical notation on Wikipedia 7 Mistake 8 Disagree with recent changes 9 [-L,L] or [-pi,pi]

Inconsistencies

I read it carefully and found several troubles.

1. In the definition we say f(x) is from L² and then, discussing convergence, we say "if f is from L² then..."

2. In the sample f(x) = x ~ ... the form is f(x) ~ a0 + ... instead of f(x) ~ a0/2 + ...

+idea: I am sure we have to emphasize the fact that the functions <math>\frac{1}{2\pi}e^{inx}<math> form an orthonormal basis and that the calculation of F_n is done by simple projecting f onto the basis functions.

Cis function and rewrite

I haven't heard of a "cis" function since I was in high school. What's wrong with "exp"? It's by far the more common notation. -- Tim Starling

I'm inclined to agree. This article has a lot of room for improvement. Maybe I'll look at it some time soon. Michael Hardy 19:50 Mar 21, 2003 (UTC)

I've done some rewriting. The "cis" function is no longer called that after my revisions. Obviously far more can be said. Maybe later I'll mention some of the vast array of applications, in physics, number theory, probability, cryptography, statistics, engineering, etc. Michael Hardy 22:17 Mar 21, 2003 (UTC)

The fourier series of an arbitrary periodic continuous function need not converge to the function.

A common error! The fourier series of an arbitrary periodic continuous function need not converge to the function.

The article does not say that the Foureir series converges to the function; the article explicitly says otherwise! I've seldom seen such sloppy reading. Michael Hardy 18:30, 27 May 2004 (UTC)

Michael,

I'm not the one who initiated this discussion, but I just took a look. I see above that you indeed understand that Fourier series of continuous functions do not converge pointwise. For the record I mention a nonconstructive proof (as you probably know, the constructive one is difficult.) Let S_nf := ∑_k=-nⁿf^(k), the value of the partial Fourier sum at 0. The space C=C(T) is a Banach space in the uniform norm; consider F:={S_n} as a family of operators on C(T). One may show that ||S_n||=||D_n||₁->∞ where D_n:=∑_k=-nⁿexp(ikx). Hence there is no uniform bound for F on C, and by the uniform boundedness principle, F is not pointwise bounded. Pick an f in C so that Ff is unbounded, in particular the Fourier series of f does not converge at 0 (not even to some value which is not f(0)). Since f is continuous, it is in particular piecewise continuous, and 0 is a point of continuity of f.

Now consider this paragraph:

That much was proved in the 19th century, as was the fact that if f is piecewise continuous then the series converges at each point of continuity.

I too find it difficult to read that particular passage correctly. Perhaps you were talking about Cesaro or Abel means? Or piecewise Holder continuous functions with exponent α>1/2? Maybe I'm misunderstanding "point of continuity."

I would be grateful if you could explain that passage to me.

Thanks,

Loisel 02:58, 29 May 2004 (UTC)

I've excised the text for the time being. I'd still like to understand what you meant, Michael. Loisel 19:53, 31 May 2004 (UTC)

Constructiveness

Since people don't seem to know, here is roughly how to construct a function (any use of Banach Steinhaus can be explicitized that way). Take a series of n_k 's going to infinity very fast. Take continuous functions f_k with <math>||f_k||\leq 1<math> such that

<math>\int f_k D_{n_k} > \frac{1}{2}||D_{n_k}||<math>

Basically, f_k should be <math>\mathrm{sign} D_{n_k}<math>, but smoothed a little so that it would be continuous. Then take

<math>f=\sum_{k=1}^\infty 2^{-k} f_k<math>

If the n_k increase sufficiently fast (you need to construct them inductively) then you get

<math>\int f D_{n_k} > \frac{1}{2}\int 2^{-k}f_k D_{n_k} > 2^{-k-2}||D_{n_k}|| \to \infty <math>

And that's it.Gadykozma 08:02, 21 Jul 2004 (UTC)

---

How about some example?

Generalized to interval a-b

I have reverted the edits by 142.150.160.187 because there are just too many errors and inconsistencies. The definition is wrong, because one of the exponents must be negative, and I believe the product of the normalizing constants must be 1/p not 2/p. The orthogonality relationships are wrong, again, one of the exponents must be negative, and the normalizing constant must be 1/p. The period of the original was 2π, but now there are at least three or four different periods, variously b-a, p, L, and 2π. I have no problem generalizing the Fourier series to an arbitrary interval a-b, and I'm not famous for error free edits myself, but lets do it 90% right at least. Paul Reiser 20:45, 17 Mar 2005 (UTC)

on an interval [a,b), period is b-a, the normalizing constant is 2/(b-a), for example on interval [-Pi, Pi), the normalizing constant is 2/(Pi+Pi) = 1/Pi. Or can be written as 1/L on an interval [-L, L). If the interval is [0, 2*Pi), the constant would be 2/(2*pi-0). which is 1/Pi. So 2/p for a period p, and 1/L for interval [-L, L) are consistent statements. You are right about orthogonality, i made a typo.

What is the relationship between p, a, and b? That needs to be up front. At present, the orthogonality in the continuous space is

<math>\frac{2}{p}\int_a^b e^{\frac{2\pi inx}{p}}e^{-\frac{2\pi imx}{p}}dx=\delta_{mn}<math>

That means when m=n we must have

<math>\frac{2}{p}\int_a^b dx=1<math>

which means

<math>\frac{2(b-a)}{p}=1<math>

which means

<math>p=2(b-a)<math>

Is this the relationship between p and a and b? If it is, then the definitions of the Fourier series and its inverse are inconsistent. If we use the definition of F_n as it is now, and substitue the definition of f(x) as it is now, from the definition section, then we have:

<math>f(x')=\sum_{n=-\infty}^\infty \left[\frac{2}{p}\int_a^bf(x)e^{\frac{-2\pi inx}{p}}dx\right] e^{\frac{2\pi inx'}{p}}<math>

<math>=\frac{2}{p}\int_a^bf(x)\sum_{n=-\infty}^\infty e^{\frac{-2\pi in(x'-x)}{p}}dx <math>

Using the orthogonality in the continuous space, the sum in the above equation is equal to

<math>\sum_{n=-\infty}^\infty \delta\left(\frac{2\pi}{p}(x-x')+2\pi n\right)<math>

which has a spike whenever x-x'=p. If we use p=2(b-a), then that happens twice in the interval b-a and we have

<math>f(x')=2f(x')<math>

which is wrong.

Another problem is that we want to have the same period used in everything. The orthogonality relationships are used to prove the validity of the inverse, the convolution theorems, Plancherel and Pareseval (as above). We should not be using one period (p) for the definition and orthogonality in the discrete space, another period (2 π) for orthogonality in the continuous space and yet another (L) in the convolution theorems. They may be correct but it makes a mess of the article. PAR 17:09, 18 Mar 2005 (UTC)

To the anonimous contributor: Please make yourself an account. That would be much helpful. Thanks. Oleg Alexandrov 00:50, 19 Mar 2005 (UTC)

How to compose good-looking non-TeX mathematical notation on Wikipedia

(See below. Michael Hardy 22:25, 17 Mar 2005 (UTC))

period is b-a,

period is b − a,

on an interval [a,b),

on an interval [a, b),

the normalizing constant is 2/(b-a),

the normalizing constant is 2/(b − a),

the normalizing constant is 2/(Pi+Pi) = 1/Pi.

the normalizing constant is 2/(π + π) = 1/π.

interval is [0, 2*Pi), the

interval is [0, 2π), the

would be 2/(2*pi-0).

would be 2/(2π − 0).

So 2/p for a period p,

So 2/p for a period p,

[-L, L)

[−L, L)

Mistake

I've calculated the continuous orthogonality relationship for a general interval [a,b]. For [-π,π] the orthogonality relationship is:

<math>\frac{1}{2\pi}\int_{-\pi}^\pi e^{inx}e^{-imx}\,dx = \delta_{nm}<math>

Assume p=b-a and make the substitution

<math>y=a+\left(\frac{x}{2\pi}+\frac{1}{2}\right)p<math>

so that y=[a,b] for x=[-π,π]. This gives for the orthogonality relationship:

<math>\frac{1}{p}e^{2\pi i(m-n)(a/p+1/2)}

\int_{a}^b e^{2\pi iny/p}e^{-2\pi imy/p}\,dy = \delta_{mn}<math>

I haven't checked this, but that exponential mess before the integral will probably be showing up in the definitions, Plancherel, convolutions, etc. etc. so we need to make it disappear by using x=[-L,L] where L is anything, including possibly π. PAR 20:24, 22 Mar 2005 (UTC)

Disagree with recent changes

This page now is quite inconsistent, as it deals with Fourier series on [a, b], then on [-π, π], and then on [-L, L]. I suggest putting things back to [-π, π], as the generalization from there to an arbitrary interval is trivial. Other ideas? Oleg Alexandrov 05:23, 20 Mar 2005 (UTC)

I agree, this is silly. It should just use [-π, π], [-L, L], or [-L/2, L/2], which are the most common conventions. In any case, a single convention should be used throughout. (At most, you can have a single subsection at the end that explains how to switch to different periods and intervals.) —Steven G. Johnson 05:47, Mar 20, 2005 (UTC)

Its silly and worse - its wrong. Look at the second orthogonality relationship. Set m=n and very quickly you get 2=1. I tried to explain this to this person a few days ago, but it doesn't seem to sink in, and any repair gets re-damaged without discussion. PAR 17:31, 20 Mar 2005 (UTC)

Certainly, standardise to period 2π at the outset. Charles Matthews 11:13, 20 Mar 2005 (UTC)

I asked Chubby Chicken (the recent contributor) to take a look here at this conversation. Let us not rush to revert things for now, but in a couple of days we definitly need to bring this article in good shape. Oleg Alexandrov 20:11, 20 Mar 2005 (UTC)

fixed mistake, the normalizing constant is 1/(b-a) for a compelx fourier series, 2/(b-a) for a real fourier series (User:Chubby Chicken forgot to sign).

But you are avoiding the bigger issue of the inconsisties you introduced with your changes (see above). And there is no gain in doing things on [a, b] rather than the original [-π, π]. Oleg Alexandrov 04:09, 21 Mar 2005 (UTC)

Well, there might be a small gain in a list of mathematical formulae. On the other hand it is usual in an encyclopedia article to simplify in minor ways, for clarity. We should do that, here. Charles Matthews 09:16, 21 Mar 2005 (UTC)

the gain is it describes fourier series expansion for function defined on any intgerval [a,b], instead of restricting to functions that are defined on [-pi pi]. (User:Chubby Chicken forgot to sign).

As I have just said, the greater generality goes against normal practice. Charles Matthews

But the article is inconsistent now, see what I wrote above. It will be easier to do everything on [-pi, pi], and then only at the end put what you wrote as generalization.

Also, please read the Fourier series article from beginning to end. Don't focus just on one section. Oleg Alexandrov 18:59, 21 Mar 2005 (UTC)

[-L,L] or [-pi,pi]

It seems to me we now have to decide whether to keep the [-pi,pi] interval or go to the [-L,L] interval. The [a,b] interval, when done correctly, is an abomination (please see Mistake above). I'm in favor of the [-L,L] interval. It simply replaces pi with L in all the formulas, no sweat. Maybe Chubby Chicken's desire to generalize is not totally misplaced. I know some are in favor of [-pi,pi]. Can we get a consensus somehow? I would be glad to write the page with [-L,L] if thats what we decide. PAR 20:24, 22 Mar 2005 (UTC)

[-L, L] is fine with me. Oleg Alexandrov 20:50, 22 Mar 2005 (UTC)

I'm in favor of [-pi, pi] here. Working with [-L, L] is going to throw a factor pi/L into the picture if I'm not mistaken; it's a slight complication but I really don't see a need for making this topic any more obscure than it needs to be. A section which describes how to go from [-pi, pi] to [-L, L] or [a, b] is OK by me but that shouldn't be front and center. For what it's worth, Wile E. Heresiarch 21:37, 22 Mar 2005 (UTC)

Now I would lean to agree with Wile. We can put some formulas for [-L, L] at the end. And the [a, b] case needs no treatment at all. Anybody knowing at least calculus will figure out how to do the shift from [-L, L]. Oleg Alexandrov 23:51, 22 Mar 2005 (UTC)

Wile is right, exp(inx) becomes exp(iπnx/L) so I'm less enthusiastic about [-L,L]. I still like [-L,L] because its easier to substitute L=π into the equations than to make the substitution x=yπ/L in the equations, including the dx and then realize that y=xL/π changes the limits of the integrals to [-L,L]. PAR 00:29, 23 Mar 2005 (UTC)