Talk:Fourier series

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)

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)Reply

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)Reply

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)Reply

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 $||f_{k}||\leq 1$ such that

\int f_{k}D_{n_{k}}>{\frac {1}{2}}||D_{n_{k}}||

Basically, f_k should be $\mathrm {sign} D_{n_{k}}$ , but smoothed a little so that it would be continuous. Then take

f=\sum _{k=1}^{\infty }2^{-k}f_{k}

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

\int fD_{n_{k}}>{\frac {1}{2}}\int 2^{-k}f_{k}D_{n_{k}}>2^{-k-2}||D_{n_{k}}||\to \infty

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 1/p for a period p, and 2/L for interval [-L, L) are consistent statements. You are right about orthogonality, i made a typo.