Re: Fourier Series
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sat, 01 Dec 2007 06:18:42 -0600
On 30 Nov 2007 19:02:33 GMT, grubb@xxxxxxxxxxxxxxxxx (Daniel Grubb)
wrote:
A better reason to withdraw the second question is that it
simply makes no sense! If a function is not continuous
it doesn't _have_ a Fourier series, so a question about
the Fourier series of a possibly non-perodic function
just doesn't make sense.
Actually, this is not true. It is perfectly possible to talk
about the Fourier series of an almost periodic function. Such
functions do not need to be periodic,
Well, yes and no. Yes, almost periodic functions are
often said to have Fourier series. No, I don't think
that this is the most common meaning of the term.
In fact one might say that it's functions on compact
abelian groups that have Fourier series (the classical
case being the circle, and the case you mention
corresponding to the Bohr compactification of R.)
nor do their Fourier
series have to converge to the original function.
An answer to the original (second) question is
f(x)=\sum_n=1^infty (1/n!) sin(x/n)
The Fourier series for this function is exactly what you
would think it should be and it converges uniformly to
f(x) (clearly!). It is also fairly claer that f(x) is not
periodic (the periods of the partial sums are unbounded).
--Dan Grubb
************************
David C. Ullrich
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