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Re: Integrating scattered data
From: |
Jordi Gutiérrez Hermoso |
Subject: |
Re: Integrating scattered data |
Date: |
Tue, 21 Aug 2007 14:17:57 -0500 |
On 21/08/07, Thomas Shores <address@hidden> wrote:
> On Monday 20 August 2007 11:27:04 am Jordi Gutiérrez Hermoso wrote:
> > I have a surface in some irregular domain of R^2 that I'm sampling
> > at scattered, unstructured points. I'd like to find the volume
> > under this surface.
> >
> > Ideally, I'd like to not interpolate to points not in my data set,
> > since my surface is rather wiggly and I don't think interpolation
> > could be very accurate. A Delaunay triangulation is an obvious
> > first step, and then I could compose my surface of triangles to
> > get some sort of O(h) integration.
> >
> > Can I do better? Are there routines already in Octave to help me
> > do this? Is it possible to do some analogue of Simpson's rule to
> > get O(h^2) precision?
>
> Unless there is additional information about the data that you are
> not providing, it seems to me that deterministic methods don't make
> a whole lot of sense in this context.
It's a fairly regular surface whose values I only know at a few
scattered points, where "few" is of the order of 1e3. It oscillates
quite a bit, but it's supposed to be smooth, as it's supposed to be
the solution of a wave equation.
> You mention O(h), which I presume is something like the maximum
> distance between sample domain points.
"Linear" was my intent. Sure, h is the maximum distance between the
points. Well, the maximum of the minimum taken for each point.
> If your sampling is really unstructured, then you have little
> control over h,
It's not that unstructured. I can control h. It just doesn't fall in
general in any sort of nice grid. A Delaunay triangulation looks fine
and more or less regular, but I can't do much better than that. Some
triangles are slim, some are fat.
> usual order estimates (which are premised on h and factors such as
> reasonably small higher derivatives and regularity of the domain,
> which you suggest you may not have)
By regularity of the domain I didn't mean that the boundary isn't
Lipschitz. I merely meant that it isn't something as nice in general
as a box or a circle. We can assume the boundary is parametrised by
some C^\infty curve, though.
> Is the domain of function generating the surface even completely
> known?
Yes, but I'm approximating this surface by scattering points inside
this domain on its boundary in an unstructured manner.
> Moreover, by use of the word "sampling" you suggest that there might
> be error in your sampled values. Is this so?
No. Perhaps it was a bad choice of words. What I meant to say is that
I don't know the surface's values at all points, only on my scattered
points.
> I'm assuming that what you are after the integral of a function
> f(x,y) over a recognizable bounded domain D in R^2.
Yes.
> What would make more sense is to me in this situation is a Monte
> Carlo method.
Ick. Can't do better?
> These have the disadvantages of relatively low accuracy and
> probabilistic estimates of error to boot,
And slow!
> but they have advantages of being relatively less sensitive to
> higher dimensional integration and ease of implementation.
Are two dimensions really that high dimensional? I was hoping I could
work with some kind of spline surface and integrate that spline
surface instead, which is what I meant when I spoke of a Simpson's
rule analogue.
> Do a google on something like "multivariate Monte Carlo methods" and
> you'll get a lot of information.
Let's call this Plan B. ;-)
I don't think my problem is ugly enough to warrant Monte Carlo
methods. Or do you disagree?
- Jordi G. H.
Re: Integrating scattered data, Thomas Shores, 2007/08/21
- Re: Integrating scattered data,
Jordi Gutiérrez Hermoso <=