Gibbs Phenomenon

On Jun 27, 2:49 am, Thomas Richter <t...@math.tu-berlin.de> wrote:
> Dave schrieb:
>
> > On Jun 26, 3:52 am, "vasindagi" <vish...@gmail.com> wrote:
> >> Hi,
> >> I m trying to learn more about gibbs phenomenon. I found various sites
> >> explaining about what is gibbs phenomenon but none of them epxlained the
> >> reason behind the phenomenon.
> >> Can anyone please explain the reason for the gibbs phenomenon or send me
> >> links about the same.
> >> Thanks in advance.
>
> > They explanation about the limitation of not using higher frequencies
> > isn't really correct - even though it displays the same type of
> > behaviour. Even if you used an infinite number of frequencies the
> > result does not converge at  that point i.e. the discontinuity.
>
> It doesn't converge *pointwise* at the discontinuity, but it does
> converge in l^2 sense, and it also converges pointwise at every epsilon
>
> > 0 around the discontinuity.
> > So what you're seeing is the limitation of sum of sinusoids
> > representation - it does not form a complete basis.
>
> Whether there are complete basis of function spaces is a matter of
> believe (namely if you thrust the axiom of choice). For all practical
> matters, Schauder bases like the free oscillations (cos/sin or exp(ix))
> are good enough. IOW, you can approximate any function as close as you
> wish with sinoids in the l^2 sense. They do not form a Schauder basis in
> l^\infty sense of the piecewise continuous functions, that's true, but a
> different statement.

The fourier  series doesn't converge at  a the point  of the
discontinuity and there is always a finite error, and it cannot be
made infinitely small - atleast from what I remember.

I agree with you in mostcases the representation is sufficient.

> > It is like trying
> > to represent 3D space with only 2 vectors.
>
> No, not at all. In 3D space, you will always have a positive error for
> arbitrary vectors. Here, you can make the error infinitely small.


Since the sin/cos bases doesn't form a basis for discontinuous
functions my analogy still stands. To represent the discontinuous
functions you start getting into the theory of distributions.

Cheers,
David

>Hi,
>I m trying to learn more about gibbs phenomenon. I found various sites
>explaining about what is gibbs phenomenon but none of them epxlained the
>reason behind the phenomenon. 
>Can anyone please explain the reason for the gibbs phenomenon or send me
>links about the same.
>Thanks in advance.
>

Here's some reading on the Gibbs effect.
Steve

http://www.dspguide.com/ch11/4.htm

P.S. download the pdf version from the page; the HTML graphic isn't
detailed enough. 

Dave wrote:

>> Whether there are complete basis of function spaces is a matter of
>> believe (namely if you thrust the axiom of choice). For all practical
>> matters, Schauder bases like the free oscillations (cos/sin or exp(ix))
>> are good enough. IOW, you can approximate any function as close as you
>> wish with sinoids in the l^2 sense. They do not form a Schauder basis in
>> l^\infty sense of the piecewise continuous functions, that's true, but a
>> different statement.
> 
> The fourier  series doesn't converge at  a the point  of the
> discontinuity and there is always a finite error, and it cannot be
> made infinitely small - atleast from what I remember.

Look, "convergence" requires a norm, otherwise this word makes no sense. 
And yes, indeed, the cos/sin series *do* converge to any (non-continuous 
or continuous) function *in L^2 sense*. I never claimed that they do
converge in a *pointwise* sense, which is something entirely different.

> I agree with you in mostcases the representation is sufficient.

This is not a matter of "what happens in most cases", it is really a 
matter of picking the mathematically suitable function space. (-:

>>> It is like trying
>>> to represent 3D space with only 2 vectors.
>> No, not at all. In 3D space, you will always have a positive error for
>> arbitrary vectors. Here, you can make the error infinitely small.
> 
> 
> Since the sin/cos bases doesn't form a basis for discontinuous
> functions my analogy still stands. To represent the discontinuous
> functions you start getting into the theory of distributions.

I afraid you don't understand - they *do* form a (Schauder-) basis of 
L^2, provably. But this only implies convergence in L^2 (mean-square 
error) sense, and not in pointwise sense. If you say "basis", you also 
need to define "norm". Actually, elements in L^2 are *not* functions, 
but equivalence classes of them where, specifically, function values 
*at* a point do not make sense, so to say that it "does not converge 
pointwise" makes little sense in an L^2 space.

All this sounds like mathematical nit-picking, but it is important to 
use a proper definition of words specifically in the 
infinite-dimensional cases or you get lost easily.

So long,
Thomas

On Jun 28, 4:41 pm, Thomas Richter <t...@math.tu-berlin.de> wrote:

> I afraid you don't understand - they *do* form a (Schauder-) basis of
> L^2, provably. But this only implies convergence in L^2 (mean-square
> error) sense, and not in pointwise sense. If you say "basis", you also
> need to define "norm". Actually, elements in L^2 are *not* functions,
> but equivalence classes of them where, specifically, function values
> *at* a point do not make sense, so to say that it "does not converge
> pointwise" makes little sense in an L^2 space.
>
> All this sounds like mathematical nit-picking, but it is important to
> use a proper definition of words specifically in the
> infinite-dimensional cases or you get lost easily.

I will admit that this area is not my forte, so I am a bit out of my
depth - so take what I say with a grain of salt. I agree with
everything you've said., but  I specifically didn't mention a norm.
All I meant to do was point out that it doesn't converge at the one
particular point i.e. the discontinuity.

I'm sure there must exist some  norm for which the fourier transform
doesn't form a basis. Perhaps the infinity norm? Perhaps you can tell
me?

Cheers,
Dave

"Dave" <dspguy2@netscape.net> wrote in message 
news:d656b993-fb08-4a41-8806-0bfa0978e5b2@d77g2000hsb.googlegroups.com...
> On Jun 28, 4:41 pm, Thomas Richter <t...@math.tu-berlin.de> wrote:
>
>> I afraid you don't understand - they *do* form a (Schauder-) basis of
>> L^2, provably. But this only implies convergence in L^2 (mean-square
>> error) sense, and not in pointwise sense. If you say "basis", you also
>> need to define "norm". Actually, elements in L^2 are *not* functions,
>> but equivalence classes of them where, specifically, function values
>> *at* a point do not make sense, so to say that it "does not converge
>> pointwise" makes little sense in an L^2 space.
>>
>> All this sounds like mathematical nit-picking, but it is important to
>> use a proper definition of words specifically in the
>> infinite-dimensional cases or you get lost easily.
>
> I will admit that this area is not my forte, so I am a bit out of my
> depth - so take what I say with a grain of salt. I agree with
> everything you've said., but  I specifically didn't mention a norm.
> All I meant to do was point out that it doesn't converge at the one
> particular point i.e. the discontinuity.
>
> I'm sure there must exist some  norm for which the fourier transform
> doesn't form a basis. Perhaps the infinity norm? Perhaps you can tell
> me?
>
> Cheers,
> Dave

It's the "basis set" that "forms the basis".  I don't know if the norm 
determines.  To me it's always been about whether you can construct and 
uniquely solve a set of equations on a set of points using that basis.  The 
objective of the equations / the norm / may not be important.  But I'd not 
want to comment on L3, L4, etc.

For example, you can use the Fourier cos/sin basis along with the infinity 
norm - that's what the Remez exchange algorithm does or can do as in the 
Parks-McClellan implementation... well, at least the cos basis.

So, the Fourier Transform yields an L2 approximation and the Remez exchange 
and L(inf) approximation, etc.

Fred