Re: 3D interpolating
- To: mathgroup at smc.vnet.net
- Subject: [mg109119] Re: 3D interpolating
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Mon, 19 Apr 2010 04:06:31 -0400 (EDT)
On 4/12/10 at 11:01 PM, ibmichuco at hotmail.com (michuco) wrote:
>Hi all,
>I am trying to fit a series of 3D points, say
>data = {{{0, 0}, 0}, {{1, 1}, 1}, {{2, 2}, 2}};
>with
>f = Interpolation[data]
>which gives me the error
>Interpolation::indim: The coordinates do not lie on a structured \
>tensor product grid.
>I think that I got the data format wrong, but I can't tell from the
>Help library of Mathematica 6.
>Ideas? Thanks in advance,
The following which has the same structure but more points works:
In[17]:= data = Flatten[Table[{{x, y}, LCM[x, y]}, {x, 4}, {y,
4}], 1];
f = Interpolation[data]
Out[18]= InterpolatingFunction[{{1,4},{1,4}},<>]
By default Interpolation does a cubic interpolation. Three data
points simply isn't enough to uniquely define the parameters of
a cubic curve.